__Arithmetic__

An archive of questions and answers that may be of interest to puzzle enthusiasts.
__Question 1 - arithmetic:__

A customer at a 7-11 store selected four items to buy, and was told
that the cost was $7.11. He was curious that the cost was the same
as the store name, so he inquired as to how the figure was derived.
The clerk said that he had simply multiplied the prices of the four
individual items. The customer protested that the four prices
should have been ADDED, not MULTIPLIED. The clerk said that that
was OK with him, but, the result was still the same: exactly $7.11.

What were the prices of the four items? Show Answer

The prices are: $1.20, $1.25, $1.50, and $3.16$7.11 is not the only number which works. Here are the first 160 such numbers, preceded by a count of distinct solutions for that price. Note that $7.11 has a single, unique solution.

There are plenty of solutions for five summands. Here are a few:

1 - $6.44 1 - $7.83 2 - $9.20 3 - $10.89 1 - $6.51 1 - $7.86 1 - $9.23 1 - $10.95 1 - $6.60 3 - $7.92 1 - $9.24 2 - $11.00 1 - $6.63 1 - $8.00 1 - $9.27 1 - $11.07 1 - $6.65 1 - $8.01 2 - $9.35 1 - $11.13 1 - $6.72 1 - $8.03 3 - $9.36 1 - $11.16 2 - $6.75 5 - $8.10 1 - $9.38 1 - $11.22 1 - $6.78 1 - $8.12 5 - $9.45 2 - $11.25 1 - $6.80 1 - $8.16 2 - $9.48 2 - $11.27 2 - $6.84 2 - $8.19 1 - $9.54 1 - $11.30 1 - $6.86 1 - $8.22 1 - $9.57 1 - $11.36 1 - $6.89 1 - $8.25 1 - $9.59 1 - $11.40 2 - $6.93 3 - $8.28 2 - $9.60 2 - $11.43 1 - $7.02 3 - $8.33 1 - $9.62 2 - $11.52 1 - $7.05 1 - $8.36 2 - $9.63 2 - $11.55 2 - $7.07 1 - $8.37 1 - $9.66 2 - $11.61 1 - $7.08 2 - $8.40 1 - $9.68 1 - $11.69 1 - $7.11 1 - $8.45 2 - $9.69 1 - $11.70 1 - $7.13 2 - $8.46 1 - $9.78 1 - $11.88 2 - $7.14 1 - $8.52 2 - $9.80 1 - $11.90 3 - $7.20 5 - $8.55 1 - $9.81 1 - $11.99 1 - $7.25 1 - $8.60 1 - $9.87 1 - $12.06 1 - $7.26 4 - $8.64 4 - $9.90 1 - $12.15 2 - $7.28 1 - $8.67 1 - $9.92 1 - $12.18 1 - $7.29 1 - $8.69 2 - $9.99 1 - $12.24 3 - $7.35 1 - $8.73 1 - $10.01 1 - $12.30 1 - $7.37 2 - $8.75 1 - $10.05 1 - $12.32 1 - $7.47 1 - $8.76 2 - $10.08 1 - $12.35 1 - $7.50 1 - $8.78 1 - $10.17 2 - $12.42 1 - $7.52 5 - $8.82 1 - $10.20 1 - $12.51 4 - $7.56 1 - $8.85 2 - $10.26 1 - $12.65 1 - $7.62 1 - $8.88 3 - $10.29 2 - $12.69 4 - $7.65 2 - $8.91 3 - $10.35 1 - $12.75 1 - $7.67 1 - $8.94 2 - $10.44 1 - $12.92 2 - $7.70 1 - $8.96 1 - $10.53 1 - $12.96 3 - $7.74 3 - $9.00 1 - $10.56 1 - $13.23 1 - $7.77 1 - $9.02 1 - $10.64 1 - $13.41 1 - $7.79 2 - $9.03 2 - $10.71 1 - $13.56 2 - $7.80 1 - $9.12 3 - $10.80 1 - $14.49 1 - $7.82 2 - $9.18 1 - $10.85 1 - $15.18 $8.28 -- at least two solutions

$8.47 -- at least two solutions

$8.82 -- at least two solutionsMark Johnson

mark@microunity.com ph:(408) 734-8100

There may be many approximate solutions, for example: $1.01, $1.15, $2.41, and $2.54. These sum to $7.11 but the product is 7.1100061.

__Question 2 - arithmetic.progression:__Is there an arithmetic progression of 20 or more primes? Show Answer

There is an arithmetic progression of 21 primes: 142072321123 + 1419763024680 i, 0 <= i < 21.It was discovered on 30 November 1990, by programs running in the background on a network of Sun 3 workstations in the Department of Computer Science, University of Queensland, Australia.

See: Andrew Moran and Paul Pritchard, The design of a background job on a local area network, Procs. 14th Australian Computer Science Conference, 1991, to appear.

__Question 3 - day.of.week:__

It's restful sitting in Tom's cosy den, talking quietly and sipping
a glass of his Madeira.

I was there one Sunday and we had the usual business of his clock. When the radio gave the time at the hour, the Ormolu antique was exactly 3 minutes slow.

"It loses 7 minutes every hour", my old friend told me, as he had done so many times before. "No more and no less, but I've gotten used to it that way."

When I spent a second evening with him later that same month, I remarked on the fact that the clock was dead right by radio time at the hour. It was rather late in the evening, but Tom assured me that his treasure had not been adjusted nor fixed since my last visit.

What day of the week was the second visit?

From "Mathematical Diversions" by Hunter + Madachy Show Answer

The answer is 17 days and 3 hours later, which would have been a Wednesday. This is the only other time in the same month when the two would agree at all.In 17 days the slow clock loses 17*24*7 minutes = 2856 minutes, or 47 hours and 36 minutes. In 3 hours more it loses 21 minutes, so it has lost a total of 47 hours and 57 minutes. Modulo 12 hours, it has *gained* 3 minutes so as to make up the 3 minutes it was slow on Sunday. It is now (fortnight plus 3 days) exactly accurate.

Since the clock was not adjusted since the last visit, it's also possible that the radio time shifted by one hour due to a change to or from summer daylight saving time. However, it turns out that the only additional possibilities that need to be considered are those of 4 days 15 hours later, when the clock would have lost 12 hours 57 minutes, and 29 days 15 hours later, when the clock would have lost 3 days 10 hours 57 minutes. Without even considering the rules for when in the month the clock is changed, these possible solutions are ruled out because we know that both visits were in the evening ("I spent a second evening with him"). and they involve times in a different part of the day.

__Question 4 - clock/palindromic:__

How many times per day does a digital clock display a palindromic number?
Show Answer

The problem is underspecified. Digital clocks may run from(a) 1:00 to 12:59

(b) 01:00 to 12:59

(c) 0:00 to 23:59

(d) 00:00 to 23:59

(e-h) any of the above with seconds appended, :00 to :59.I agree that not all of these are common, but I have seen some uncommon ones. For that matter, I've seen ones not on the above list -- the Toronto subway stations used to contain digital clocks that ran from 0:00 to 12:59 (pm), then from 1:00 (pm) to 11:59 (pm), while a computer that I used to use had the inverse anomaly -- its time of day command displayed times from 12:00:00 to 12:59:59 (am), then 01:00:00 to 23:59:59!

I get the following results for the 8 cases.

CASE AM+PM HOURS pals/hr TOTAL OVERALL (a) yes 1-9 6 54 10-12 1 3 114 (b) yes 01-05 1 5 06-09 0 0 10-12 1 3 16 (c) no 0-9 6 60 10-15 1 6 16-19 0 0 20-23 1 4 70 (d) no 00-05 1 6 06-09 0 0 10-15 1 6 16-19 0 0 20-23 1 4 16 (e) yes 1-9 60 540 10-12 6 18 1116 (f) yes 01-05 6 30 06-09 0 0 10-12 6 18 96 (g) no 0-9 60 600 10-15 6 36 16-19 0 0 20-23 6 24 660 (h) no 00-05 6 36 06-09 0 0 10-15 6 36 16-19 0 0 20-23 6 24 96

--Mark Brader (msb@sq.com)

__Question 5 - clock/reversible:__

How many times per day can the hour and minute hands on an analog clock switch
roles and still signify a valid time, ignoring the second hand?
Show Answer

Have 12 clocks C1, C2 ... C12 show 1:00, 2:00, ..., 12:00. Have a clock C0 show 12:00Now turn C0 around 12 hours, simultaneously turning C1-C12 so their hour hands always coincide with the minute hand of C0, i.e., as C0 spans 12 hours, C1-C12 will span 1 hour, but for each possible placing of the hour hand, all 12 possible 'true' placings of the minute hand will be represented by one of the 12 clocks.

Each time the hour hand of C0 coincides with the minute hand of a C1-C12 clock we have a reversible valid time. This happens regularly 12 times each C0 hour, but the first and last time is equal (12:00), so the number of reversible true times is 12*12-1 = 143 spaced regularly in the 12-hour interval, ie. each 5 min 2.0979+ sec

stein.kulseth@tf.tele.no [X.400] stein.kulseth@nta.no [internet]

__Question 6 - clock/right.angle:__How many times per day do the hour and minute hands of a clock form a right angle? Show Answer

44. Twice each hour equals 48, less one between 2:00 and 4:00 and one between 8:00 and 10:00 for both A.M. and P.M.

__Question 7 - thirds:__

Do the 3 hands on a clock ever divide the face of the clock into 3
equal segments, i.e. 120 degrees between each hand?
Show Answer

First let us assume that our clock has 60 divisions. We will show that any time the hour hand and the minute hand are 20 divisions (120 degrees) apart, the second hand cannot be an integral number of divisions from the other hands, unless it is straight up (on the minute).

Let us use h for hours, m for minutes, and s for seconds. We will use =n to mean congruent mod n, thus 12 =5 7.

We know that m =60 12h, that is, the minute hand moves 12 times as fast as the hour hand, and wraps around at 60. We also have s =60 60m. This simplifies to s/60 =1 m, which goes to s/60 = frac(m) (assuming s is in the range 0 <= s < 60), which goes to s = 60 frac(m). Thus, if m is 5.5, s is 30.

Now let us assume the minute hand is 20 divisions ahead of the hour hand. So m =60 h + 20, thus 12h =60 h + 20, 11h =60 20, and, finally, h =60/11 20/11 (read 'h is congruent mod 60/11 to 20/11'). So all values of m are k + n/11 for some integral k and integral n, 0 <= n < 11. s is therefore 60n/11. If s is to be an integral number of units from m and h, we must have 60n =11 n. But 60 and 11 are relatively prime, so this holds only for n = 0. But if n = 0, m is integral, so s is 0.

Now assume, instead, that the minute hand is 20 divisions behind the hour hand. So m =60 h - 20, 12h =60 h - 20, 11h =60 -20, h =60/11 -20/11. So m is still k + n/11. Thus s must be 0.

But if s is 0, h must be 20 or 40. But this translates to 4 o'clock or 8 o'clock, at both of which the minute hand is at 0, along with the second hand.

Thus the 3 hands can never be 120 degrees apart, Q.E.D.

This assumes, of course, that the second hand is synchronized with the minute hand. This is not true on some inexpensive analog watches. This also assumes that the watch is not broken (:^)).

__Question 8 - consecutive.composites:__

Are there 10,000 consecutive non-prime numbers?
Show Answer

9973!+2 through 9973!+10006 are composite.

__Question 9 - consecutive.product:__

Prove that the product of three or more consecutive positive integers cannot
be a perfect square.
Show Answer

Three consecutive numbers:

If a and b are relatively prime, and ab is a square, then a and b are squares. (This is left as an exercise.)Suppose (n - 1)n(n + 1) = k^2, where n > 1. Then n(n^2 - 1) = k^2. But n and (n^2 - 1) are relatively prime. Therefore n^2 - 1 is a perfect square, which is a contradiction.

Four consecutive numbers:

n(n + 1)(n + 2)(n + 3) = (n^2 + 3n + 1)^2 - 1Five consecutive numbers:

Assume the product is a integer square, call it m.

The prime factorization of m must have even numbers of each prime factor.For each prime factor, p, of m, p >= 5, p^2k must divide one of the consecutive naturals in the product. (Otherwise, the difference between two of the naturals in the product would be a positive multiple of a prime >= 5. But in this problem, the greatest difference is 4.) So we need only consider the primes 2 and 3.

Each of the consecutive naturals is one of:

1) a perfect square

2) 2 times a perfect square

3) 3 times a perfect square

4) 6 times a perfect square.By the shoe box principle, two of the five consecutive numbers must fall into the same category.

If there are two perfect squares, then their difference being less than five limits their values to be 1 and 4. (0 is not a natural number, so 0 and 1 and 0 and 4 cannot be the perfect squares.) But 1*2*3*4*5=120!=x*x where x is an integer.

If there are two numbers that are 2 times a perfect square, then their difference being less than five implies that the perfect squares (which are multiplied by 2) are less than 3 apart, and no two natural squares differ by only 1 or 2.

A similar argument holds for two numbers which are 3 times a perfect square.

We cannot have the case that two of the 5 consecutive numbers are multiples (much less square multiples) of 6, since their difference would be >= 6, and our span of five consecutive numbers is only 4.

Therefore the assumption that m is a perfect square does not hold.

QED.

In general the equation:

y^2 = x(x+1)(x+2)...(x+n), n > 3has only the solution corresponding to y = 0.

This is a theorem of Rigge [O. Rigge, ``Uber ein diophantisches Problem'', IX Skan. Math. Kong. Helsingfors (1938)] and Erdos [P. Erdos, ``Note on products of consecutive integers,'' J. London Math. Soc. #14 (1939), pages 194-198].

A proof can be found on page 276 of [L. Mordell, ``Diophantine Equations'', Academic Press 1969].

__Question 10 - consecutive.sums:__Find all series of consecutive positive integers whose sum is exactly 10,000. Show Answer

Generalize to find X (and I) such that

(X + X+1 + X+2 + ... + X+I) = T for any integer T.You are asking for all (X,I) s.t. (2X+I)(I+1) = 2T. The problem is (very) slightly easier if we don't restrict X to being positive, so we'll solve this first.

Note that 2X+I and I+1 must have different parities, so the answer to the relaxed question is N = 2*(o_1+1)*(o_2+1)*...*(o_n+1), where 2T = 2^o_0*3^o_1*...*p_n^o_n (the prime factorization); this is easily seen to be the number of ways we can break 2T up into two positive factors of differing parity (with order).

In particular, 20000 = 2^5*5^4, hence there are 2*(4+1) = 10 solutions for T = 10000. These are (2X+I,I+1):

(32*1,5^4) (32*5,5^3) (32*5^2,5^2) (32*5^3,5) (32*5^4,1) (5^4,32*1) (5^3,32*5) (5^2,32*5^2) (5,32*5^3) (1,32*5^4) And they give rise to the solutions (X,I):

(-296,624) (28,124) (388,24) (1998,4) (10000,0) (297,31) (-27,179) (-387,799) (-1997,3999) (-9999,19999)

If you require that X>0 note that this is true iff 2X+I > I+1 and hence the number of solutions to this problem is N/2 (due to the symmetry of the above ordered pairs).

__Question 11 - conway:__

Describe the sequence a(1)=a(2)=1, a(n) = a(a(n-1)) + a(n-a(n-1)) for n>2.
Show Answer

This sequence and its remarkable properties were discovered, apparently independently, by Douglas Hofstadter (circa 1975), John Conway (in the early 1980's), B.W. Connoly, and others. Since Conway discovered many of the deeper properties, and is the one responsible for popularizing the sequence, it seems appropriate to name the sequence after him.Some properties of a(n):

-- a(n+1) - a(n) = 0 or 1

-- a(2^n) = 2^(n-1)

-- n/2 <= a(n) <= n

-- A(n)= 2a(n) - n has zeros at the powers of 2 and positive "humps" in between

-- A(2^n + t) = A(2^(n+1) - t) for t between 0 and 2^n, i.e. the humps are symmetric

-- a(2n) <= 2a(n)

-- a(n)/n --> 1/2 as n --> infinity

a(n) and A(n) are self-similar, in the sense that their values on the (N+1)'st hump can be obtained by a "magnification" process applied to the values on the N'th hump. They are *not* chaotic sequences, since their asymptotics and combinatorial structure are very regular and rigid.

In a lecture at Bell Labs in 1988, Conway asked for the rate at which a(n)/n converges to 1/2. In particular, he asked for N(epsilon), the last value of n such that |a(n)/n - 1/2| exceeds epsilon, in the case epsilon=1/20. This problem was solved by Colin Mallows of Bell Labs: he found that N(1/20)=1489 and N(1/40)=6083008742. These values are reported in his article in the American Mathematical Monthly, January 1991, p. 5.

Conway's sequence has a deep combinatorial structure, and is intimately connected with all of the following: variants of Pascal's triangle, the Gaussian distribution, combinatorial operations on finite sets, coin-pushing games, and Fibonacci and Catalan numbers. All of this is explained in my joint paper with Ravi Vakil; anyone who would like to receive a copy in LaTex format should contact me at the e-mail address listed below.

One byproduct of our work is an algorithm to determine N(epsilon) for given positive epsilon. Here are some particular values:

e Last n such that |a(n)/n - 1/2| > e 1/20 1489 (found by Mallows in 1988) 1/30 758765 1/40 6083008742 (found by Mallows in 1988) 1/50 809308036481621 1/60 1684539346496977501739 1/70 55738373698123373661810220400 1/80 15088841875190938484828948428612052839 1/90 127565909103887972767169084026274554426122918035 1/100 8826608001127077619581589939550531021943059906967127007025 These values were computed by the Mathematica program listed below; the algorithm is explained in our paper. To use the program, load it into Mathematica and type neps[t] for some numerical value of t (which should probably be smaller than 1/5 or so). The program will output N(t), e.g. neps[1/20] = 1489. These numbers grow very fast: N(epsilon) will be about 2^((0.138015/epsilon)^2), so N(1/1000) will have about 5735 digits. The program requires very little memory space, but its runtime appears to grow rapidly as epsilon decreases (on a Sun 3, it took about one second to find the first value listed, and several minutes to find the last).

neps[eps_] := Block[{W}, L = 1 + Floor[(0.138015/eps)^2]; e = eps*2; W = processvector[L]; While[Length[W] > 0, nmax = 1 + Last[W][[4]]; L++; W = processvector[L]]; nmax]

processvector[L_] := Block[{V}, V = startvector[L]; While[notfound[V], V = newbody[V]]; V]

startvector[L_] := Select[initialvector[L], gt]

initialvector[L_] := Table[{L, b, Binomial[L - 1, b - 1], 2^(L + 1) - Sum[Binomial[L, c], {c, b, L}]}, {b, 0, L - 1}]

c[0] = 0

c[n_] := c[n] = Sum[Binomial[2*a, a]/(a + 1), {a, 0, n - 1}]

gt[x_] := U[x] > e

U[x_] := (x[[3]] + M[x[[1]], x[[2]]])/(x[[4]] + incp[x[[1]], x[[2]]])

M[n_, n_] = -1

M[n_, k_] := Binomial[n - 1, K[n, k]] - Binomial[n - 1, k - 1] + c[K[n, k]]

K[n_, k_] := Min[k, n - k]

incp[n_, k_] := Max[M[n, k], 1]

notfound[V_] := Length[V] > 0 && Last[V][[2]] > 0 && Last[V][[1]] > Last[V][[2]]

newbody[V_] := Join[Drop[V, -1], newtail[V]]

newtail[V_] := Select[{vleft[Last[V]], vright[Last[V]]}, gt]

vleft[x_] := {x[[1]] - 1, x[[2]] - 1, x[[3]], x[[4]]}

vright[x_] := {x[[1]] - 1, x[[2]], x[[3]] + S[x[[1]] - 1, x[[2]] - 1], x[[4]] + Binomial[x[[1]] - 1, x[[2]] - 1]}

S[n_, k_] := Binomial[n - 1, k] - Binomial[n - 1, k - 1]

-Tal Kubo kubo@math.harvard.edu or kubo@zariski.harvard.edu

__Question 12 - digits/6.and.7:__

Does every number which is not divisible by 5 have a multiple whose
only digits are 6 and 7?
Show Answer

Yes. My proof follows:Claim 1: For every k, there is a k-digit number whose only digits are 6 and 7, which is divisible by 2^k.

The proof is by induction. Suppose N is a k-digit number satisfying the above condition. Then either N = 0 (mod 2^(k+1)) or N = 2^k (mod 2^(k+1)). Note that 6(10^k) = 0 (mod 2^(k+1)), and 7(10^k) = 2^k (mod 2^(k+1)). So, either 6*10^k + N or 7*10^k + N is divisible by 2^(k+1).

Claim 2: If m and 10 are relatively prime, then for any r, there is a number N whose only digits are 6 and 7 such that N = r (mod m).

Proof: Let K be the (m^2)-digit number whose only digit is 6. There is an s, 0 <= s < m, so that K + s = r (mod m). Let N = K + 10^(m - 1) + 10^(2m - 2) + . . . + 10^(sm - s). Since 10^(im - i) = 1 (mod m), N = K + s (mod m) = r (mod m). Clearly, every digit of N is either 6 or 7.

Claim 3: If n is not divisible by 5, then there is a number N whose only digits are 6 and 7, so that N is divisible by n.

Proof: We can write n = (2^k)m, with gcd(m,10)=1. Use claim 1 to find a k-digit number M, whose only digits are 6 and 7, which is divisible by 2^k. Choose an integer r so that (10^k)r + M = 0 (mod m). Use claim 2 to find a number K whose only digits are 6 and 7, so that K = r (mod m). Let N = 10^k K + M. Then N = 0 (mod m) and N = 0 (mod 2^k), so N is divisible by n. Finally, the only digits of N are 6 and 7, so we are done.

David Radcliffe radcliff@csd4.csd.uwm.edu

__Question 13 - digits/all.ones:__

Prove that some multiple of any integer ending in 3 contains all 1s.
Show Answer

Let n be our integer; one such desired multiple is then (10^(phi(n))-1)/9. All we need is that (n,10) = 1, and if the last digit is 3 this must be the case. A different proof using the pigeonhole principle is to consider the sequence 1, 11, 111, ..., (10^n - 1)/9. We must have at some point that either some member of our sequence = 0 (mod n) or else some value (mod n) is duplicated. Assume the latter, with x_a and x_b, x_b>x_a, possesing the duplicated remainders. We then have that x_b - x-a = 0 (mod n). Let m be the highest power of 10 dividing x_b - x_a. Now since (10,n) = 1, we can divide by 10^m and get that (x_b - x_a)/10^m = 0 (n). But (x_b - x_a)/10^m is a number containing only the digit 1.

Q.E.D.

__Question 14 - digits/arabian:__

What is the Arabian Nights factorial, the number x such that x! has 1001
digits? How about the prime x such that x! has exactly 1001 zeroes on
the tail end. (Bonus question, what is the 'rightmost' non-zero digit in x!?)
Show Answer

The first answer is 450!.Determining the number of zeroes at the end of x! is relatively easy once you realize that each such zero comes from a factor of 10 in the product

1 * 2 * 3 * ... * x

Each factor of 10, in turn, comes from a factor of 5 and a factor of 2. Since there are many more factors of 2 than factors of 5, the number of 5s determines the number of zeroes at the end of the factorial.

The number of 5s in the set of numbers 1 .. x (and therefore the number of zeroes at the end of x!) is:

z(x) = int(x/5) + int(x/25) + int(x/125) + int(x/625) + ...

This series terminates when the powers of 5 exceed x.

I know of no simple way to invert the above formula (i.e., to find x for a given z(x)), but I can approximate it by noting that, except for the "int" function,

5*z(x) - x = z(x)

which gives:

x = 4*z(x) (approximately).

The given problem asked, "For what prime x is z(x)=1001". By the above forumla, this is approximately 4*1001=4004. However, 4004! has only

800 + 160 + 32 + 6 + 1 = 999 zeroes at the end of it.

The numbers 4005! through 4009! all have 1000 zeroes at their end and the numbers 4010! through 4014! all have 1001 zeroes at their end.

Since the problem asked for a prime x, and 4011 = 3*7*191, the only solution is x=4013.

The problem of determining the rightmost nonzero digit in x! is somewhat more difficult. If we took the numbers 1, 2, ... , x and removed all factors of 5 (and an equal number of factors of 2), the remaining numbers multiplied together modulo 10 would be the answer. Note that since there are still many factors of 2 left, the rightmost nonzero digit must be 2, 4, 6, or 8 (x > 1).

Letting r(x) be the rightmost nonzero digit in x!, an expression for r(x) is:

r(x) = (r(int(x/5)) * w * r(x mod 10)) mod 10, x >= 10.

where w is 4 if int(x/10) is odd and 6 if it is even.

The values of r(x) for 0 <= x <= 9 are 1, 1, 2, 6, 4, 2, 2, 4, 2, and 8.

The way to see this is true is to take the numbers 1, 2, ..., x in groups of 10. In each group, remove 2 factors of 10. For example, from the set 1, 2, ..., 10, choose a factor of 2 from 2 and 6 and a factor of 5 from 5 and 10. This leaves 1, 1, 3, 4, 1, 3, 7, 8, 9, 2. Next, separate all the factors that came from multiples of 5. The rightmost nonzero digit of x! can now (hopefully) be seen to be:

r(x) = (r(int(x/5)) * w * r(x mod 10)) mod 10

where w is the rightmost digit in the number formed by multiplying the numbers 1, 2, 3, ..., 10*int(x/10) after the factors of 10 and the factors left over by the multiples of 5 have been removed. In the example with x = 10, this would be (1 * 1 * 3 * 4 * 3 * 7 * 8 * 9) mod 10 = 4. The "r(x mod 10)" term takes care of the numbers from 10*int(x/10)+1 up to x.

The "w" term can be seen to be 4 or 6 depending on whether int(x/10) is odd or even since, after removing 10*n+5 and 10*n+10 and a factor of 2 each from 10*n+2 and 10*n+6 from the group of numbers 10*n+1 through 10*n+10, the remaining factors (mod 10) always equals 4 and 4^t mod 10 = 4 if t is odd and 6 when t is even (t != 0).

So, finally, the rightmost nonzero digit in 4013! is found as follows:

r(4013) = (r(802) * 4 * 6) mod 10

r(802) = (r(160) * 6 * 2) mod 10

r(160) = (r(32) * 6 * 1) mod 10

r(32) = (r(6) * 4 * 2) mod 10Using a table of r(x) for 0 <= x <= 9, r(6) = 2. Then,

r(32) = (2 * 4 * 2) mod 10 = 6

r(160) = (6 * 6 * 1) mod 10 = 6

r(802) = (6 * 6 * 2) mod 10 = 2

r(4013) = (2 * 4 * 6) mod 10 = 8

Thus, the rightmost nonzero digit in 4013! is 8.

__Question 15 - digits/circular:__

What 6 digit number, with 6 different digits, when multiplied by all integers
up to 6, circulates its digits through all 6 possible positions, as follows:

ABCDEF * 1 = ABCDEF

ABCDEF * 3 = BCDEFA

ABCDEF * 2 = CDEFAB

ABCDEF * 6 = DEFABC

ABCDEF * 4 = EFABCD

ABCDEF * 5 = FABCDE
Show Answer

ABCDEF=142857 (the digits of the expansion of 1/7).

__Question 16 - digits/divisible:__

Find the least number using 0-9 exactly once that is evenly divisible by each
of these digits.
Show Answer

Since the sum of the digits is 45, any permutation of the digits gives a multiple of 9. To get a multiple of both 2 and 5, the last digit must be 0, and thus to get a multiple of 8 (and 4), the tens digit must be even, and the hundreds digit must be odd if the tens digit is 2 or 6, and even otherwise. The number will also be divisible by 6, since it is divisible by 2 and 3, so 7 is all we need to check. First, we will look for a number whose first five digits are 12345; now, 1234500000 has a remainder of 6 when divided by 7, so we have to arrange the remaining digits to get a remainder of 1. The possible arrangements, in increasing order, are78960, remainder 0

79680, remainder 6

87960, remainder 5

89760, remainder 6

97680, remainder 2

98760, remainder 4That didn't work, so try numbers starting with 12346; this is impossible because the tens digit must be 8, and the hundreds digit cannot be even. Now try 12347, and 1234700000 has remainder 2. The last five digits can be

58960, remainder 6

59680, remainder 5, so this works, and the number is

1234759680.

__Question 17 - digits/equations/123456789:__

In how many ways can "." be replaced with "+", "-", or "" (concatenate) in
.1.2.3.4.5.6.7.8.9=1 to form a correct equation?
Show Answer

1-2 3+4 5+6 7-8 9 = 1

+1-2 3+4 5+6 7-8 9 = 1

1+2 3+4-5+6 7-8 9 = 1

+1+2 3+4-5+6 7-8 9 = 1

-1+2 3-4+5+6 7-8 9 = 1

1+2 3-4 5-6 7+8 9 = 1

+1+2 3-4 5-6 7+8 9 = 1

1-2 3-4+5-6 7+8 9 = 1

+1-2 3-4+5-6 7+8 9 = 1

1-2-3-4 5+6 7-8-9 = 1

+1-2-3-4 5+6 7-8-9 = 1

1+2-3 4+5 6-7-8-9 = 1

+1+2-3 4+5 6-7-8-9 = 1

-1+2 3+4+5-6-7-8-9 = 1

-1 2+3 4-5-6+7-8-9 = 1

1+2+3+4-5+6+7-8-9 = 1

+1+2+3+4-5+6+7-8-9 = 1

-1+2+3-4+5+6+7-8-9 = 1

1-2-3+4+5+6+7-8-9 = 1

+1-2-3+4+5+6+7-8-9 = 1

1+2 3+4 5-6 7+8-9 = 1

+1+2 3+4 5-6 7+8-9 = 1

1+2 3-4-5-6-7+8-9 = 1

+1+2 3-4-5-6-7+8-9 = 1

1+2+3+4+5-6-7+8-9 = 1

+1+2+3+4+5-6-7+8-9 = 1

-1+2+3+4-5+6-7+8-9 = 1

1-2+3-4+5+6-7+8-9 = 1

+1-2+3-4+5+6-7+8-9 = 1

-1-2-3+4+5+6-7+8-9 = 1

1-2+3+4-5-6+7+8-9 = 1

+1-2+3+4-5-6+7+8-9 = 1

1+2-3-4+5-6+7+8-9 = 1

+1+2-3-4+5-6+7+8-9 = 1

-1-2+3-4+5-6+7+8-9 = 1

-1+2-3-4-5+6+7+8-9 = 1

-1+2 3+4 5-6 7-8+9 = 1

1-2 3-4 5+6 7-8+9 = 1

+1-2 3-4 5+6 7-8+9 = 1

-1+2 3-4-5-6-7-8+9 = 1

-1+2+3+4+5-6-7-8+9 = 1

1-2+3+4-5+6-7-8+9 = 1

+1-2+3+4-5+6-7-8+9 = 1

1+2-3-4+5+6-7-8+9 = 1

+1+2-3-4+5+6-7-8+9 = 1

-1-2+3-4+5+6-7-8+9 = 1

1+2-3+4-5-6+7-8+9 = 1

+1+2-3+4-5-6+7-8+9 = 1

-1-2+3+4-5-6+7-8+9 = 1

-1+2-3-4+5-6+7-8+9 = 1

1-2-3-4-5+6+7-8+9 = 1

+1-2-3-4-5+6+7-8+9 = 1

1-2 3+4+5+6+7-8+9 = 1

+1-2 3+4+5+6+7-8+9 = 1

1+2+3+4 5-6 7+8+9 = 1

+1+2+3+4 5-6 7+8+9 = 1

1 2+3 4+5-6 7+8+9 = 1

+1 2+3 4+5-6 7+8+9 = 1

1+2+3-4-5-6-7+8+9 = 1

+1+2+3-4-5-6-7+8+9 = 1

-1+2-3+4-5-6-7+8+9 = 1

1-2-3-4+5-6-7+8+9 = 1

+1-2-3-4+5-6-7+8+9 = 1

-1-2-3-4-5+6-7+8+9 = 1

-1-2 3+4+5+6-7+8+9 = 1

1-2+3 4-5 6+7+8+9 = 1

+1-2+3 4-5 6+7+8+9 = 1

1 2-3 4+5-6+7+8+9 = 1

+1 2-3 4+5-6+7+8+9 = 1Total solutions = 69 (26 of which have a leading "+", which is redundant)

69/19683 = 0.35 %

for those who care (it's not very elegant but it did the trick):

#include <stdio.h>

#include <math.h>

main (argc,argv)

int argc;

char *argv[];

{

int sresult, result, operator[10],thisop;

char buf[256],ops[3];

int i,j,tot=0,temp;

ops[0] = ' ';

ops[1] = '-';

ops[2] = '+';

for (i=1; i<10; i++) operator[i] = 0;

for (j=0; j < 19683; j++) {

result = 0;

sresult = 0;

thisop = 1;

for (i=1; i<10; i++) {

switch (operator[i]) {

case 0:

sresult = sresult * 10 + i;

break;

case 1:

result = result + sresult * thisop;

sresult = i;

thisop = -1;case 2:

result = result + sresult * thisop;

sresult = i;

thisop = 1;

break;

}

}

result = result + sresult * thisop;

if (result == 1) {

tot++;

for (i=1;i<10;i++)

printf("%c%d",ops[operator[i]],i);

printf(" = %d\n",result);

}

temp = 0;

operator[1] += 1;

for (i=1;i<10;i++) {

operator[i] += temp;

if (operator[i] > 2) { operator[i] = 0; temp = 1;}

else temp = 0;

}

}

printf("Total solutions = %d\n" , tot);

}

cwren@media.mit.edu (Christopher Wren)

__Question 18 - digits/equations/1992:__

1 = -1+9-9+2. Extend this list to 2 through 100 on the left side of
the equals sign.
Show Answer

1 = -1+9-9+2

2 = 1*9-9+2

3 = 1+9-9+2

4 = 1+9/9+2

5 = 1+9-sqrt(9)-2

6 = 1^9+sqrt(9)+2

7 = -1+sqrt(9)+sqrt(9)+2

8 = 19-9-2

9 = (1/9)*9^2

10= 1+(9+9)/2

11= 1+9+sqrt(9)-2

12= 19-9+2

13= (1+sqrt(9))!-9-2

14= 1+9+sqrt(9)!-2

15= -1+9+9-2

16= -1+9+sqrt(9)!+2

17= 1+9+9-2

18= 1+9+sqrt(9)!+2

19= -1+9+9+2

20= (19-9)*2

21= 1+9+9+2

22= (-1+sqrt(9))*(9-2)

23= (1+sqrt(9))!-sqrt(9)+2

24= -1+9*sqrt(9)-2

25= 1*9*sqrt(9)-2

26= 19+9-2

27= 1*9+9*2

28= 1+9+9*2

29= 1*9*sqrt(9)+2

30= 19+9+2

31= (1+sqrt(9))!+9-2

32= -1+sqrt(9)*(9+2)

33= 1*sqrt(9)*(9+2)

34= (-1+9+9)*2

35= -1+(9+9)*2

36= 1^9*sqrt(9)!^2

37= 19+9*2

38= 1*sqrt(9)!*sqrt(9)!+2

39= 1+sqrt(9)!*sqrt(9)!+2

40= (1+sqrt(9)!)*sqrt(9)!-2

41= -1+sqrt(9)!*(9-2)

42= (1+sqrt(9))!+9*2

43= 1+sqrt(9)!*(9-2)

44= -1+9*(sqrt(9)+2)

45= 1*9*(sqrt(9)+2)

46= 1+9*(sqrt(9)+2)

47= (-1+sqrt(9)!)*9+2

48= 1*sqrt(9)!*(sqrt(9)!+2)

49= (1+sqrt(9)!)*(9-2)

50= (-1+9)*sqrt(9)!+2

51= -1+9*sqrt(9)!-2

52= 1*9*sqrt(9)!-2

53= -1+9*sqrt(9)*2

54= 1*9*sqrt(9)*2

55= 1+9*sqrt(9)*2

56= 1*9*sqrt(9)!+2

57= 1+9*sqrt(9)!+2

58= (1+9)*sqrt(9)!-2

59= 19*sqrt(9)+2

60= (1+9)*sqrt(9)*2

61= (1+sqrt(9)!)*9-2

62= -1+9*(9-2)

63= 1*9*(9-2)

64= 1+9*(9-2)

65= (1+sqrt(9)!)*9+2

66= 1*sqrt(9)!*(9+2)

67= 1+sqrt(9)!*(9+2)

68= -(1+sqrt(9))!+92

69= (1+sqrt(9))!+(9/.2)

70= (1+9)*(9-2)

71= -1-9+9^2

72= (1+sqrt(9))*9*2

73= -19+92

74= (-1+9)*9+2

75= -1*sqrt(9)!+9^2

76= 1-sqrt(9)!+9^2

77= (1+sqrt(9)!)*(9+2)

78= -1+9*9-2

79= 1*9*9-2

80= 1+9*9-2

81= 1*9*sqrt(9)^2

82= -1+9*9+2

83= 1*9*9+2

84= 1+9*9+2

85= -1-sqrt(9)!+92

86= -1*sqrt(9)!+92

87= 1-sqrt(9)!+92

88= (1+9)*9-2

89= -1*sqrt(9)+92

90= 1-sqrt(9)+92

91= -1^9+92

92= (1+9)*9+2

93= 1^9+92

94= -1+sqrt(9)+92

95= 19*(sqrt(9)+2)

96= -1+99-2

97= 1*99-2

98= 1+99-2

99= 1*9*(9+2)

100= -1+99+2

__Question 19 - digits/equations/24:__

Form an expression that evaluates to 24 that contains two 3's, two 7's,
and zero or more of the operators +, -, *, and /, and parentheses. What
about two 4's and two 7's, or three 5's and one 1, or two 3's and two 8's?
Show Answer

7*(3+3/7)

7*(4-4/7)

5*(5-1/5)

8/(3-8/3)

__Question 20 - digits/equations/383:__Make 383 out of 1,2,25,50,75,100 using +,-,*,/. Show Answer

You can get 383 with ((2+50)/25+1)*100+75.Of course, if you expect / as in C, the above expression is just 375. But then you can get 383 with (25*50-100)/(1+2). Pity there's no way to use the 75.

If we had a language that rounded instead of truncating, we could use ((1+75+100)*50)/(25-2) or (2*75*(25+100))/(50-1).

I imagine your problem lies in not dividing things that aren't divisible.

Dan Hoey

Hoey@AIC.NRL.Navy.Mil

__Question 21 - digits/equations/find:__

Write a program for finding expressions built out of given numbers and using
given operators that evaluate to a given value, or listing all possible values.
Show Answer

As set up, it requires recompilation for different sets of numbers; it's currently set up for 8,8,3,3; to try in other numbers, stick 'em in the 'val' array. To find all target numbers for which the equation is valid, uncomment the 't' loop and 'target = t', and extend the range to be checked... you might want to turn off VERBOSE. I don't bother with eliminating symmetries if equal vals are given (like 8 8 3 3), so I normally use it likenumop 24 | sort | uniq

As it stands, this gives the output:

8 / (3 - (8 / 3)) = 24.0

8 / (3 - (8 / 3)) = 24.0

8 / (3 - (8 / 3)) = 24.0

8 / (3 - (8 / 3)) = 24.0As you can see, there are five different kinds of binary trees with exactly four leaf nodes. The program tries all four operators in each place, and all four values in each of the leaves, guaranteeing that each is used only once... a fairly quick operation. A small extract from 'numop 1' shows the five different shapes of trees:

((3 * 8) / 3) / 8 = 1.0

(3 * (8 / 3)) / 8 = 1.0

(3 - 3) + (8 / 8) = 1.0

3 * ((8 / 3) / 8) = 1.0

3 * (8 / (3 * 8)) = 1.0Probably FUDGE ought to be set a little lower, for more confidence that the equality isn't fortuitous. Extensions to other binary operators are obvious; unary operators and more values are not. For a more general problem I'd go recursive, use exact rational arithmetic, and have a fine old time.

Jim Gillogly

21 Wedmath S.R. 1993, 10:58/* numop: using elementary operations on 4 numbers, find a

* desired result; e.g. 24.

*

* Don't worry about symmetries resulting in multiple correct answers.

*

* 11 Aug 93, SCRYER

*/

#include <stdio.h>

#define VERBOSE

#define MUL 0

#define DIV 1

#define ADD 2

#define SUB 3

#define FUDGE 0.01

float val[4] = {8, 8, 3, 3};

float eval(), atof(), fabs();

char nameop();

int divzero;

main(argc, argv)

int argc;

char *argv[];

{

int op1, op2, op3;

int iv1, iv2, iv3, iv4;

int used[4];

int i;

float target;

float e1, e2, e3;

int t, winner;

if (argc != 2)

{

fprintf(stderr, "Usage: numop\n");

exit(1);

}

target = atof(argv[1]);

/* for (t = -1000; t < 1000; t++) */

{

/* target = t;*/

winner = 0;

for (i = 0; i < 4; i++) used[i] = 0;

for (op1 = 0; op1 < 4; op1++)for (op2 = 0; op2 < 4; op2++)for (op3 = 0; op3 < 4; op3++)for (iv1 = 0; iv1 < 4; iv1++)

{

used[iv1] = 1;

for (iv2 = 0; iv2 < 4; iv2++)

{

if (used[iv2]) continue;

used[iv2] = 1;

for (iv3 = 0; iv3 < 4; iv3++)

{

if (used[iv3]) continue;

used[iv3] = 1;

for (iv4 = 0; iv4 < 4; iv4++)

{

if (used[iv4]) continue;

/* Case 1 */

divzero = 0;

e3 = eval(op3, val[iv3], val[iv4]);

e2 = eval(op2, val[iv1], val[iv2]);

e1 = eval(op1, e2, e3); /* (u + v) * (w - x) */

if (fabs(e1 - target) < FUDGE && ! divzero)

#ifdef VERBOSE

printf("(%.0f %c %.0f) %c (%.0f %c %.0f) = %.1f\n",

val[iv1], nameop(op2), val[iv2], nameop(op1),

val[iv3], nameop(op3), val[iv4], e1);

#else

winner = 1; #endif

/* Case 2 */

divzero = 0;

e3 = eval(op3, val[iv1], val[iv2]);

e2 = eval(op2, e3, val[iv3]);

e1 = eval(op1, e2, val[iv4]); /* ((u + v) * w) - x */

if (fabs(e1 - target) < FUDGE && ! divzero)

#ifdef VERBOSE

printf("((%.0f %c %.0f) %c %.0f) %c %.0f = %.1f\n",

val[iv1], nameop(op3), val[iv2], nameop(op2), val[iv3], nameop(op1), val[iv4], e1);

#else

winner = 1;

#endif

/* Case 3 */

divzero = 0;

e3 = eval(op3, val[iv2], val[iv3]);

e2 = eval(op2, val[iv1], e3);

e1 = eval(op1, e2, val[iv4]); /* (u + (v * w)) - x */

if (fabs(e1 - target) < FUDGE && ! divzero)

#ifdef VERBOSE

printf("(%.0f %c (%.0f %c %.0f)) %c %.0f = %.1f\n",

val[iv1], nameop(op2), val[iv2], nameop(op3), val[iv3],

nameop(op1), val[iv4], e1);

#else

winner = 1;

#endif

/* Case 4 */

divzero = 0;

e3 = eval(op3, val[iv2], val[iv3]);

e2 = eval(op2, e3, val[iv4]);

e1 = eval(op1, val[iv1], e2); /* u + ((v * w) - x) */

if (fabs(e1 - target) < FUDGE && ! divzero)

#ifdef VERBOSE

printf("%.0f %c ((%.0f %c %.0f) %c %.0f) = %.1f\n",

val[iv1], nameop(op1), val[iv2], nameop(op3), val[iv3],

nameop(op2), val[iv4], e1);

#else

winner = 1;

#endif

/* Case 5 */ /* u + (v * (w - x)) */

divzero = 0;

e3 = eval(op3, val[iv3], val[iv4]);

e2 = eval(op2, val[iv2], e3);

e1 = eval(op1, val[iv1], e2);

if (fabs(e1 - target) < FUDGE && ! divzero)

#ifdef VERBOSE

printf("%.0f %c (%.0f %c (%.0f %c %.0f)) = %.1f\n",

val[iv1], nameop(op1), val[iv2], nameop(op2), val[iv3],

nameop(op3), val[iv4], e1);

#else

winner = 1;

#endif

}

used[iv3] = 0;

}

used[iv2] = 0;

}

}

#ifndef VERBOSE

if (winner) printf("%d\n", t), fflush(stdout);

#endif

}

}

char nameop(op)

int op;

{

switch(op)

{

case MUL: return '*';

case DIV: return '/';

case ADD: return '+';

case SUB: return '-';

}

return '?';

}

float eval(op, val1, val2)

int op;

float val1, val2;

{

switch(op)

{

case MUL: return val1 * val2;

case DIV:

if (val2 == 0.)

{

divzero = 1;

#ifdef EXTREMELYVERBOSE

fprintf(stderr, "Division by zero.\n");

#endif

}

return val2 == 0.? 0. : val1 / val2;

case ADD: return val1 + val2;

case SUB: return val1 - val2;

}

return 0.;

}

}

char nameop(op)

int op;

{

switch(op)

{

case MUL: return '*';

case DIV: return '/';

case ADD: return '+';

case SUB: return '-';

}

return '?';

}

float eval(op, val1, val2)

int op;

float val1, val2;

{

switch(op)

{

case MUL: return val1 * val2;

case DIV:

if (val2 == 0.)

{

divzero = 1;

#ifdef EXTREMELYVERBOSE

fprintf(stderr, "Division by zero.\n");

#endif

}

return val2 == 0.? 0. : val1 / val2;

case ADD: return val1 + val2;

case SUB: return val1 - val2;

}

return 0.;

}

__Question 22 - digits/extreme.products:__

What are the extremal products of three three-digit numbers using digits 1-9?
Show Answer

There is a simple procedure which applies to these types of problems (and which can be proven with help from the arithmetic-geometric inequality).For the first part we use the "first large then equal" procedure. This means that are three numbers will be 7**, 8**, and 9**. Now the digits 4,5,6 get distributed so as to make our three number as close to each other as possible, i.e. 76*, 85*, 94*. The same goes for the remaining three digits, and we get 763, 852, 941.

For the second part we use the "first small then different" procedure. Our three numbers will be of the form 1**, 2**, 3**. We now place the three digits so as to make our three numbers as unequal as possible; this gives 14*, 25*, 36*. Finishing, we get 147, 258, 369.

Now, *prove* that these procedures work for generalizations of this problem.

** Question 23 - digits/labels:**
You have an arbitrary number of model kits (which you assemble for
fun and profit). Each kit comes with twenty (20) stickers, two of which
are labeled "0", two are labeled "1", ..., two are labeled "9".
You decide to stick a serial number on each model you assemble starting
with one. What is the first number you cannot stick. You may stockpile
unused numbers on already assembled models, but you may not crack open
a new model to get at its stickers. You complete assembling the current
model before starting the next.
Show Answer

The method I used for this problem involved first coming up with a formula that says how many times a digit has been used in all n models.n = k*10^i + m for some k,m with 0

f(d,n) = (number of d's used getting to k*10^i from digits 0 to (i-1)) + (number of d's used by #'s 10^i to n from digit i) + f(d,m) f(d,n) = i*k*10^(i-1) + (if (d < k) 10^i else if (d == k) m+1 else 0) + f(d,m)

This doesn't count 0's, which should be ok as they are not used as often as other digits. From the formula, it is clear that f(1,n) is never less than f(d,n) for 1So I just calculated f(1,n) for various n (with some help from bc).

I quickly discovered that for n = 2*10^15, f(1,n) = 2*n. After further trials I determined that for n = 1999919999999981, f(1,n) = 2*n + 1. This appears to be the smallest n with f(1,n) > 2*n.

__Question 24 - digits/least.significant/factorial:__

What is the least significant non-zero digit in the decimal expansion of n!?
Show Answer

Reduce mod 10 the numbers 2..n and then cancel out the required factors of 10. The final step then involves computing 2^i*3^j*7^k mod 10 for suitable i,j and k.A small program that performs this calculation is appended. Like the other solutions, it takes O(log n) arithmetic operations.

-kym

===

#include

#include

int p[6][4]={

/*2*/ 2, 4, 8, 6,

/*3*/ 3, 9, 7, 1,

/*4*/ 4, 6, 4, 6,

/*5*/ 5, 5, 5, 5,

/*6*/ 6, 6, 6, 6,

/*7*/ 7, 9, 3, 1,

};

main(){

int i;

int n;

for(n=2;n<1000;n++){

i=lsdfact(n);

printf("%d\n",i);

}

exit(0);

}

lsdfact(n){

int a[10];

int i;

int n5;

int tmp;

for(i=0;i<=9;i++)a[i]=alpha(i,n);

n5=0;

/* NOTE: order is important in following */

l5:;

while(tmp=a[5]){ /* cancel factors of 5 */

n5+=tmp;

a[1]+=(tmp+4)/5;

a[3]+=(tmp+3)/5;

a[5]=(tmp+2)/5;

a[7]+=(tmp+1)/5;

a[9]+=(tmp+0)/5;

}

l10:;

if(tmp=a[0]){

a[0]=0; /* cancel all factors of 10 */

for(i=0;i<=9;i++)a[i]+=alpha(i,tmp);

}

if(a[5]) goto l5;

if(a[0]) goto l10;

/* n5 == number of 5's cancelled;

must now cancel same number of factors of 2 */

i=ipow(2,a[2]+2*a[4]+a[6]+3*a[8]-n5)*

ipow(3,a[3]+a[6]+2*a[9])*

ipow(7,a[7]);

assert(i%10); /* must not be zero */

return i%10;

}

alpha(d,n){

/* number of decimal numbers in [1,n] ending in digit d */

int tmp;

tmp=(n+10-d)/10;

if(d==0)tmp--; /* forget 0 */

return tmp;

}

ipow(x,y){

/* x^y mod 10 */

if(y==0) return 1;

if(y==1) return x;

return p[x-2][(y-1)%4];

}

__Question 25 - digits/least.significant/tower.of.power:__

What are the least significant digits of 9^(8^(7^(6^(5^(4^(3^(2^1))))))) ?
Show Answer

9^11 = 9 (mod 100), so we need to find 8^...^1 (mod 10).

8^5 = 8 (mod 10), so we need to find 7^...^1 (mod 4).

7^3 = 7 (mod 4), so we need to find 6^...^1 (mod 2), but

this is clearly 0, so 7^...^1 = 1 (mod 4) ==>

8^...^1 = 8 (mod 10) ==> 9^...^1 = 9^8 (mod 100) = 21 (mod 100).

** Question 26 - digits/most.significant/googol:**
What digits does googol! start with?
Show Answer

I'm not sure how to calculate the first googol of digits of log10(e), but here's the first 150(approximately) of them...0.43429448190325182765112891891660508229439700580366656611445378316586464920

8870774729224949338431748318706106744766303733641679287158963906569221064663We need to deal with the digits immediately after the decimal point in googol*log10(e), which are .187061

frac[log(googol!)] = frac[halflog2pi + 50 + googol(100-log10(e))]

= frac{halflog2pi + frac[googol(100-log10(e))]}

= frac[.399090 + (1- .187061)]

= .21202910 ** .212029 = 1.629405

Which means that googol! starts with 1629

__Question 27 - digits/most.significant/powers:__

What is the probability that 2^N begins with the digits 603245?
Show Answer

2^N begins with 603245 iff 603246*10^m > 2^N >= 603245*10^m for some positive integer m ==> m+log(603246) > N*log(2) >= m+log(603245); so 2^N begins with 603245 iff frac(log(603246)) > frac(N*log(2)) >= frac(log(603245)). If we are using natural density then N*log(2) is uniformly distributed mod 1 since log(2) is irrational, hence the probability is frac(log(603246)) - frac(log(603245)) = frac(log(603246)-log(603245)) = frac(log(603246/603245)).

A neat observation is that since it is known p_n*c, where p_n is the nth prime and c is irrational, is uniformly distributed mod 1, we get the same answer if we replace 2^N with 2^{p_n}.

--

Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618

__Question 28 - digits/nine.digits:__

Form a number using 0-9 once with its first n digits divisible by n.
Show Answer

First, reduce the sample set. For each digit of ABCDEFGHI, such that the last digit, (current digit), is the same as a multiple of N :A: Any number 1-9

B: Even numbers 2,4,6,8 (divisible by 2).

C: Any number 1-9 (21,12,3,24,15,6,27,18,9).

D: Even numbers 2,4,6,8 (divisible by 4, every other even).

E: 5 (divisible by 5 and 0 not allowed).

F: Even numbers (12,24,6,18)

G: Any number 1-9 (21,42,63,14,35,56,7,28,49).

H: Even numbers (32,24,16,8)

I: Any number 1-9 (81,72,63,54,45,36,27,18,9)Since E must be 5, I can eliminate it everywhere else.

Since I will use up all the even digits, (2,4,6,8) filling in those spots that must be even. Any number becomes all odds, except 5.

A: 1,3,7,9

B: 2,4,6,8

C: 1,3,7,9

D: 2,4,6,8

E: 5

F: 2,4,6,8

G: 1,3,7,9

H: 2,4,6,8

I: 1,3,7,9We have that 2C+D=0 (mod 4), and since C is odd, this implies that D is 2 or 6; similarly we find that H is 2 or 6 ==> {B,F} = {4,8}. D+5+F=0 (mod 3) ==> if D=2 then F=8, if D=6 then F=4.

We have two cases.

Assume our number is of the form A4C258G6I0. Now the case n=8 ==> G=1,9; case n=3 ==> A+1+C=0 (mod 3) ==> {A,C}={1,7} ==> G=9, I=3. The two numbers remaining fail for n=7.

Assume our number is of the form A8C654G2I0. The case n=8 ==> G=3,7. If G=3, we need to check to see which of 1896543, 9816543, 7896543, and 9876543 are divisible by 7; none are.

If G=7, we need to check to see which of 1896547, 9816547, 1836547, and 3816547 are divisible by 7; only the last one is, which yields the solution 3816547290.

__Question 29 - digits/palindrome:__

Does the series formed by adding a number to its reversal always end in
a palindrome?
Show Answer

This is not known.If you start with 196, after 9480000 iterations you get a 3924257-digit non-palindromic number. However, there is no known proof that you will never get a palindrome.

The statement is provably false for binary numbers. Roland Sprague has shown that 10110 starts a series that never goes palindromic.

** Question 30 - digits/palintiples:**
Find all numbers that are multiples of their reversals.
Show Answer

We are asked to find numbers that are integer multiples of their reversals, which I call palintiples. Of course, all the palindromic numbers are a trivial example, but if we disregard the unit multiples, the field is narrowed considerably.Rouse Ball (_Mathematical_recreations_and_essays_) originated the problem, and G. H. Hardy (_A_mathematician's_apology_) used the result that 9801 and 8712 are the only four-digit palintiples as an example of a theorem that is not ``serious''. Martin Beech (_The_mathema- tical_gazette_, Vol 74, #467, pp 50-51, March '90) observed that 989*01 and 879*12 are palintiples, an observation he ``confirmed'' on a hand calculator, and conjectured that these are all that exist.

I confirm that Beech's numbers are palintiples, I will show that they are not all of the palintiples. I will show that the palintiples do not form a regular language. And then I will prove that I have found all the palintiples, by describing the them with a generalized form of regular expression. The results become more interesting in other bases.

First, I have a more reasonable method of confirming that these numbers are palintiples:

Proof: First, letting "9*" and "0*" refer an arbitrary string of nines and a string of zeroes of the same length, I note that

879*12 = 879*00 + 12 = (880*00 - 100) + 12 = 880*00 - 88

219*78 = 219*00 + 78 = (220*00 - 100) + 78 = 220*00 - 22989*01 = 989*00 + 1 = (990*00 - 100) + 1 = 990*00 - 99

109*89 = 109*00 + 89 = (110*00 - 100) + 89 = 110*00 - 11It is obvious that 4x(220*00 - 22) = 880*00 - 88 and that

9x(110*00 - 11) = 990*00 - 99. QED.Now, to show that these palintiples are not all that exist, let us take the (infinite) language L[4] = (879*12 + 0*), and let Pal(L[4]) refer to the set of palindromes over the alphabet L[4]. It is immediate that the numbers in Pal(L[4]) are palintiples. For instance,

8712 000 87912 879999912 879999912 87912 000 8712

= 4 x 2178 000 21978 219999978 219999978 21978 000 2178

(where I have inserted spaces to enhance readability) is a palintiple. Similarly, taking L[9] = (989*01 + 0*), the numbers in Pal(L[9]) are palintiples. We exclude numbers starting with zeroes.

The reason these do not form a regular language is that the sub-palintiples on the left end of the number must be the same (in reverse order) as the sub-palintiples on the right end of the number:8712 8712 87999912 = 4 x 2178 2178 21999978

is not a palintiple, because 8712 8712 87999912 is not the reverse of 2178 2178 21999978. The pumping lemma can be used to prove that Pal(L[4])+Pal(L[9]) is not a regular language, just as in the familiar proof that the palindromes over a non-singleton alphabet do not form a regular language.

Now to characterize all the palintiples, let N be a palintiple, N=CxR(N), where R(.) signifies reversal, and C>1 is an integer. (I use "x" for multiplication, to avoid confusion with the Kleene star "*", which signifies the concatenated closure.) If D is a digit of N, let D' refer to the corresponding digit of R(N). Since N=CxR(N), D+10T = CxD'+S, where S is the carry in to the position occupied by D' when R(N) is multiplied by C, and T is the carry out of that position. Similarly, D'+10T'=CxD+S', where S', T' are carries in and out of the position occupied by D when R(N) is multiplied by C.

Since D and D' are so closely related, I will use the symbol D:D' to refer to a digit D on the left side of a string with a corresponding digit D' on the right side of the string. More formally, an expression "x[1]:y[1] x[2]:y[2] ... x[n]:y[n] w" will refer to a string "x[1] x[2] ... x[n] w y[n] ... y[2] y[1]", where the x[i] and y[i] are digits and w is a string of zero or one digits. So 989901 may be written as 9:1 8:0 9:9 and 87912 may be written as 8:2 7:1 9. Thus Pal(L[4])+Pal(L[9]) (omitting numbers with leading zeroes) can be represented as

(8:2 7:1 9:9* 1:7 2:8 0:0*)*

(0:0* + 0 + 8:2 7:1 ( 9:9* + 9:9* 9))

+ (9:1 8:0 9:9* 0:8 1:9 0:0*)*

(0:0* + 0 + 9:1 8:0 ( 9:9* + 9:9* 9)). (1)For each pair of digits D:D', there are a very limited--and often empty--set of quadruples S,T,S',T' of digits that satisfy the equations

D +10T =CxD'+S

D'+10T'=CxD +S', (2)yet such a quadruple must exist for "D:D'" to appear in a palintiple with multiplier C. Furthermore, the S and T' of one D:D' must be T and S', respectively, of the next pair of digits that appear. This enables us to construct a finite state machine to recognize those palintiples. The states [X#Y] refer to a pair of carries in D and D', and we allow a transition from state [T#S'] to state [S#T'] on input symbol D:D' exactly when equations (2) are satisfied. Special transitions for a single-digit input symbol (the central digit of odd-length palintiples) and the criteria for the initial and the accepting states are left as exercises. The finite state machines thus formed are

State Accept Symbol New State Symbol New State Symbol New State [0#0] Y 8:2 [0#3] 0:0 [0#0] 0 [A] [0#3] N 7:1 [3#3] [3#3] Y 1:7 [3#0] 9:9 [3#3] 9 [A] [3#0] N 2:8 [0#0] [A] Y for constant C=4, and

State Accept Symbol New State Symbol New State Symbol New State [0#0] Y 1:9 [0#8] 0:0 [0#0] 0 [A] [0#8] N 8:0 [8#8] [8#8] Y 0:8 [8#0] 9:9 [8#8] 9 [A] [8#0] N 9:1 [0#0] [A] Y for constant C=9, and the finite state machines for other constants accept only strings of zeroes. It is not hard to verify that the proposed regular expression (1) represents the union of the languages accepted by these machines, omitting the empty string and strings beginning with zero.

I have written a computer program that constructs finite state machines for recognizing palintiples for various bases and constants. I found that base 10 is actually an unusually boring base for this problem. For instance, the machine for base 8, constant C=5 is

State Accept Symbol New State Symbol New State Symbol New State [0#0] Y 0:0 [0#0] 5:1 [0#3] 0 [A] [0#3] N 1:0 [1#1] 6:1 [1#4] [1#1] Y 0:1 [3#0] 5:2 [3#3] [3#0] N 1:5 [0#0] 6:6 [0#3] 6 [A] [3#3] Y 2:5 [1#1] 7:6 [1#4] [1#4] N 1:1 [4#1] 6:2 [4#4] 1 [A] [4#4] Y 2:6 [4#1] 7:7 [4#4] 7 [A] [4#1] N 1:6 [3#0] 6:7 [3#3] [A] Y for which I invite masochists to write the regular expression. If anyone wants more, I should remark that the base 29 machine for constant C=18 has 71 states!

By the way, I did not find any way of predicting the size or form of the machines for the various bases, except that the machines for C=B-1 all seem to be isomorphic to each other. If anyone investigates the general behavior, I would be most happy to hear about it.

Dan Hoey

Hoey@AIC.NRL.Navy.Mil

May, 1992

[ A preliminary version of this message appeared in April, 1991. ]

Dan

__ Question 31 - digits/power.two:__

Prove that for any 9-digit number (base 10) there is an integral power
of 2 whose first 9 digits are that number.
Show Answer

Let v = log to base 10 of 2.

Then v is irrational.Let w = log to base 10 of these 9 digits.

Since v is irrational, given epsilon > 0, there exists some natural number n such that

{w} < {nv} < {w} + epsilon

({x} is the fractional part of x.) Let us pick n for when

epsilon = log 1.00000000000000000000001.

Then 2^n does the job.

__Question 32 - digits/prime/101:__

How many primes are in the sequence 101, 10101, 1010101, ...?
Show Answer

Note that the sequence

101 , 10101, 1010101, ....

can be viewed as

100**1 +1, 100**2 + 100**1 + 1, 100**3 + 100**2 + 100**1 +1 ....

that is,

the k-th term in the sequence is

100**k + 100**(k-1) + 100**(k-2) + ...+ 100**(1) + 1

= (100)**(k+1) - 1

----------------

11 * 9

= (10)**(2k+2) - 1

----------------

11 * 9

= ((10)**(k+1) - 1)*((10)**(k+1) +1)

---------------------------------

11*9

hus either 11 and 9 divide the numerator. Either they both divide the same factor in the numerator or different factors in the numerator. In any case, after dividing, they leave the numerators as a product of two integers. Only in the case of k = 1, one of the integers is 1. Thus there is exactly one prime in the above sequence: 101.

__Question 33 - digits/prime/all.prefix:__

What is the longest prime whose every proper prefix is a prime?
Show Answer

23399339, 29399999, 37337999, 59393339, 73939133

__Question 34 - digits/prime/change.one:__

What is the smallest number that cannot be made prime by changing a single
digit? Are there infinitely many such numbers?
Show Answer

200. Obviously, you would have to change the last digit, but 201, 203, 207, and 209 are all composite. For any smaller number, you can change the last digit, and get

2,11,23,31,41,53,61,71,83,97,101,113,127,131,149,151,163,173,181, or 191.

200+2310n gives an infinite family, because changing the last digit to 1 or 7 gives a number divisible by 3; to 3, a number divisible by 7; to 9, a number divisible by 11.

__Question 35 - digits/prime/prefix.one:__

2 is prime, but 12, 22, ..., 92 are not. Similarly, 5 is prime
whereas 15, 25, ..., 95 are not. What is the next prime number
which is composite when any digit is prefixed?
Show Answer

149

__Question 36 - digits/reverse:__

Is there an integer that has its digits reversed after dividing it by 2?
Show Answer

Assume there's such a positive integer x such that x/2=y and y is the reverse of x.Then x=2y. Let x = a...b, then y = b...a, and:

b...a (y)

x 2

--------

a...b (x)From the last digit b of x, we have b = 2a (mod 10), the possible values for b are 2, 4, 6, 8 and hence possible values for (a, b) are (1,2), (6,2), (2,4), (7,4), (3,6), (8,6), (4,8), (9,8).

From the first digit a of x, we have a = 2b or a = 2b+1. None of the above pairs satisfy this condition. A contradiction.

Hence there's no such integer.

__Question 37 - digits/rotate:__

Find integers where multiplying them by single digits rotates their digits
one position, so that the last digit become the first digit.
Show Answer

2 105263157894736842

3 1034482758620689655172413793

4 102564 153846 179487 205128 230769

5 142857 102040816326530612244897959183673469387755

6 1016949152542372881355932203389830508474576271186440677966

1186440677966101694915254237288135593220338983050847457627

1355932203389830508474576271186440677966101694915254237288

1525423728813559322033898305084745762711864406779661016949

7 1014492753623188405797 1159420289855072463768 1304347826086956521739

8 1012658227848 1139240506329

9 10112359550561797752808988764044943820224719In base B, suppose you have an N-digit answer A whose digits are rotated when multiplied by K. If D is the low-order digit of A, we have

(A-D)/B + D B^(N-1) = K A .

Solving this for A we have

D (B^N - 1)

A = ----------- .

B K - 1In order for A >= B^(N-1) we must have D >= K. Now we have to find N such that B^N-1 is divisible by R=(BK-1)/gcd(BK-1,D). This always has a minimal solution N0(R,B)

N0(ST,B)=N0(S,B)N0(T,B) when (S,T)=1, and for prime powers, N0(P^X,B) divides (P-1)P^(X-1). Determining which divisor is a little more complicated but well-known (cf. Hardy & Wright).

So given B and K, there is one minimal solution for each D=K,K+1,...,B-1, and you get all the solutions by taking repetitions of the minimal solutions.

__Question 38 - digits/sesqui__

Find the least number where moving the first digit to the end multiplies by 1.5.
Show Answer

Let's represent this number as a*10^n+b, where 1<=a<=9 and b < 10^n. Then the condition to be satisfied is:3/2(a*10^n+b) = 10b+a

3(a*10^n+b) = 20b+2a

3a*10^n+3b = 20b+2a

(3*10^n-2)a = 17b

b = a*(3*10^n-2)/17

So we must have 3*10^n-2 = 0 (mod 17) (since a is less than 10, it cannot contribute the needed prime 17 to the factorization of 17b). (Also, assuming large n, we must have a at most 5 so that b < 10^n will be satisfied, but note that we can choose a=1). Now,

3*10^n-2 = 0 (mod 17)

3*10^n = 2 (mod 17)

10^n = 12 (mod 17)

A quick check shows that the smallest n which satisfies this is 15 (the fact that one exists was assured to us because 17 is prime). So, setting n=15 and a=1 (obviously) gives us b=176470588235294, so the number we are looking for is

1176470588235294

and, by the way, we can set a=2 to give us the second smallest such number,

2352941176470588

Other things we can infer about these numbers is that there are 5 of them less than 10^16, 5 more less than 10^33, etc.

__Qurstion 39 - digits/squares/change.leading__

What squares remain squares when their leading digits are incremented?
Show Answer

Omitting solutions that are obtained from smaller solutions by multiplying by powers of 10, the squares of these numbers satisfy the condition:1. (105,145), (3375,4625), (14025,17225), (326625,454625), (10846875,14753125), (43708125,53948125), ...

2. (45,55), (144375,175625), (463171875,560828125), ...

7. (2824483699753370361328125,2996282391593370361328125), ...

Here is how to find them. We have (y+x)*(y-x) = 10^n, and so we must have {y+x, y-x} as {5^m*10^a, 2^m*10^b} in some order. It is also necessary (and sufficient) that y/x lies in the interval [sqrt(3/2),sqrt(2)], or equivalently that (y+x)/(y-x) lies in [3+sqrt(8),5+sqrt(24)] = [5.82842...,9.89897...]. Thus we need to make (5/2)^m*10^(a-b), or its reciprocal, in this range. For each m there is clearly at most one power of 10 that will do. m=2,a=b gives (105,145); m=3,b=a+2 gives (3375,4625), and so on.

There are infinitely many non-equivalent solutions, because log(5/2) / log(10) is irrational.

One can use exactly the same argument to find squares whose initial 2 can be replaced by a 3, of course, except that the range of (y+x)/(y-x) changes.

__Question 40 - digits/squares/length.22:__

Is it possible to form two numbers A and B from 22 digits such that
A = B^2? Of course, leading digits must be non-zero.
Show Answer

No, the number of digits of A^2 must be of the form 3n or 3n-1.

__Question 41 - digits/squares/length.9__

Is it possible to make a number and its square, using the digits from 1
through 9 exactly once?
Show Answer

Yes, there are two such pairs: (567, 567^2=321489) and (854,854^2=729316).

__Question 42 - digits/squares/three.digits__

What squares consist entirely of three digits (e.g., 1, 4, and 9)?
Show Answer

The full set of solutions up to 10**12 is1 -> 1

2 -> 4

3 -> 9

7 -> 49

12 -> 144

21 -> 441

38 -> 1444

107 -> 11449

212 -> 44944

31488 -> 9914 94144

70107 -> 49149 91449

3 87288 -> 14 99919 94944

956 10729 -> 9 14141 14499 11441

4466 53271 -> 199 49914 44949 99441

31487 17107 -> 9914 41941 99144 49449

2 10810 79479 -> 4 44411 91199 99149 11441

If the algorithm is used in the form I presented it before, generating the whole set P_n before starting on P_{n+1}, the store requirements begin to become embarassing. For n>8 I switched to a depth-first strategy, generating all the elements in P_i (i=9..12) congruent to a particular x in P_8 for each x in turn. This means the solutions don't come out in any particular order, of course. CPU time was 16.2 seconds (IBM 3084).

In article <1990Feb6.025205.28153@sun.soe.clarkson.edu>, Steven Stadnicki suggests alternate triples of digits, in particular {1,4,6} (with many solutions) and {2,4,8} (with few). I ran my program on these as well, up to 10**12 again:

1 -> 1

2 -> 4

4 -> 16

8 -> 64

12 -> 144

21 -> 441

38 -> 1444

108 -> 11664

119 -> 14161

121 -> 14641

129 -> 16641

204 -> 41616

408 -> 1 66464

804 -> 6 46416

2538 -> 64 41444

3408 -> 116 14464

6642 -> 441 16164

12908 -> 1666 16464

25771 -> 6641 44441

78196 -> 61146 14416

81619 -> 66616 61161

3 33858 -> 11 14611 64164

2040 00408 -> 41 61616 64641 66464

6681 64962 -> 446 44441 64444 61444

8131 18358 -> 661 16146 41166 16164

40182 85038 -> 16146 61464 66146 61444 (Steven's last soln.)

1 20068 50738 -> 1 44164 46464 46111 44644

1 26941 38988 -> 1 61141 16464 66616 64144

1 27069 43631 -> 1 61466 41644 14114 64161

4 01822 24262 -> 16 14611 14664 16614 44644

4 05784 63021 -> 16 46611 66114 66644 46441

78 51539 12392 -> 6164 66666 14446 44111 61664

and

2 -> 4

22 -> 484

168 -> 28224

478 -> 2 28484

2878 -> 82 82884 (Steven's last soln.)

2109 12978 -> 44 48428 42888 28484

(so the answer to Steven's "Are there any more at all?" is "Yes".)The CPU times were 42.9 seconds for {1,4,6}, 18.7 for {2,4,8}. This corresponds to an interesting point: the abundance of solutions for {1,4,6} is associated with abnormally large sets P_n (|P_8| = 16088 for {1,4,6} compared to |P_8| = 5904 for {1,4,9}) but the deficiency of solutions for {2,4,8} is *not* associated with small P_n's (|P_8| = 6816 for {2,4,8}). Can anyone wave a hand convincingly to explain why the solutions for {2,4,8} are so sparse?

I suspect we are now getting to the point where an improved algorithm is called for. The time to determine all the n-digit solutions (i.e. 2n-digit squares) using this last-significant-digit-first is essentially constant * 3**n. Dean Hickerson in <90036.134503HUL@PSUVM.BITNET>, and Ilan Vardi in <1990Feb5.214249.22811@Neon.Stanford.EDU>, suggest using a most-significant-digit-first strategy, based on the fact that the first n digits of the square determine the (integral) square root; this also has a running time constant * 3**n. Can one attack both ends at once and do better?

Chris Thompson

JANET: cet1@uk.ac.cam.phx

Internet: cet1%phx.cam.ac.uk@nsfnet-relay.ac.ukHey guys, what about

648070211589107021 ^ 2 = 419994999149149944149149944191494441This was found by David Applegate and myself (about 5 minutes on a DEC 3100, program in C).

This is the largest square less than 10^42 with the 149-property; checking took a bit more than an hour of CPU time.

As somebody suggested, we used a combined most-significant/least-significant digits attack. First we make a table of p-digit prefixes (most significant p digits) that could begin a root whose square has the 149 property in its first p digits. We organize this table into buckets by the least significant q digits of the prefixes. Then we enumerate the s digit suffixes whose squares have the 149 property in their last s digits. For each such suffix, we look in the table for those prefixes whose last q digits match the first q of the suffix. For each match, we consider the p + s - q digit number formed by overlapping the prefix and the suffix by q digits. The squares of these overlap numbers must contain all the squares with the 149 property.

The time expended is O(3^p) to generate the prefix table, O(3^s) to enumerate the suffixes, and O(3^(p+s) / 10^q) to check the overlaps (being very rough and ignoring the polynomial factors) By judiciously chosing p, q, and s, we can fix things so that each bucket of the table has around O(1) entries: set q = p log10(3). Setting p = s, we end up looking for squares whose roots have n = 2 - log10(3) digits, with an algorithm that takes time O( 3 ^ [n / (2 - log10(3)]) ), roughly time O(3^[.66n]). Compared to the O(3^n) performance of either single-ended algorithm, this lets us check 50% more digits in the same amount of time (ignoring polynomial factors). Of course, the space cost of the combined-ends method is high.

-- Guy and Dave

--

Guy Jacobson School of Computer Science

Carnegie Mellon arpanet : guy@cs.cmu.edu

Pittsburgh, PA 15213 csnet : Guy.Jacobson%a.cs.cmu.edu@csnet-relay

(412) 268-3056 uucp : ...!{seismo, ucbvax, harvard}!cs.cmu.edu!guyHere is an algorithm which takes O(sqrt(n)log(n)) steps to find all perfect squares < n whose only digits are 1, 4 and 9.

This doesn't sound too great *but* it doesn't use a lot of memory and only requires addition and <. Also, the actual run time will depend on where the first non-{1,4,9} digit appears in each square.

set n = 1

set odd = 1while(n < MAXVAL)

{

if(all digits of n are in {1,4,9})

{

print n

}add 2 to odd

add odd to n

}This works because (X+1)^2 - x^2 = 2x+1. That is, if you start with 0 and add successive odd numbers to it you get 0+1=1, 1+3=4, 4+5=9, 9+7=16 etc. I've started the algorithm at 1 for convenience.

The "O" value comes from looking at at most all digits (log(n)) of all perfect squares < n (sqrt(n) of them) at most a constant number of times.

I didn't save the articles with algorithms claiming to be O(3^log(n)) so I don't know if their calculations needed to (or did) account for multiplication or sqrt() of large numbers. O(3^log(n)) sounds reasonable so I'm going to assume they did unless I hear otherwise.

Any comments? Please email if you just want to refresh my memory on the other algorithms.

Andrew Charles

acgd@ihuxy.ATT.COMM

__Question 43 - digits/squares/twin__

Let a twin be a number formed by writing the same number twice,
for instance, 81708170 or 132132. What is the smallest square twin?
Show Answer

1322314049613223140496 = 36363636364 ^ 2.

The key to solving this puzzle is looking at the basic form of these "twin" numbers, which is some number k = 1 + 10^n multiplied by some number 10^(n-1) <= a < 10^n. If ak is a perfect square, k must have some repeated factor, since a

__Question 44 - digits/sum.of.digits:__Find sod ( sod ( sod (4444 ^ 4444 ) ) ). Show Answer

let X = 4444^4444sod(X) <= 9 * (# of digits) < 145900

sod(sod(X)) <= sod(99999) = 45

sod(sod(sod(X))) <= sod(39) = 12but sod(sod(sod(X))) = 7 (mod 9)

thus sod(sod(sod(X))) = 7

__Question 45 - digits/zeros/million:__

How many zeros occur in the numbers from 1 to 1,000,000?
Show Answer

In the numbers from 10^(n-1) through 10^n - 1, there are 9 * 10^(n-1) numbers of n digits each, so 9(n-1)10^(n-1) non-leading digits, of which one tenth, or 9(n-1)10^(n-2), are zeroes. When we change the range to 10^(n-1) + 1 through 10^n, we remove 10^(n-1) and put in 10^n, gaining one zero, sop(n) = p(n-1) + 9(n-1)10^(n-2) + 1 with p(1)=1.

Solving the recurrence yields the closed form

p(n) = n(10^(n-1)+1) - (10^n-1)/9.

For n=6, there are 488,895 zeroes, 600,001 ones, and 600,000 of all other digits.

__Question 46 - digits/zeros/trailing:__

How many trailing zeros are in the decimal expansion of n!?
Show Answer

The general answer to the question "what power of p divides x!" where p is prime

is (x-d)/(p-1) where d is the sum of the digits of (x written in base p).

So where p=5, 10 is written as 20 and is divisible by 5^2 (2 = (10-2)/4);

x to base 10: 100 1000 10000 100000 1000000 x to base 5: 400 13000 310000 11200000 224000000 d : 4 4 4 4 8 trailing 0s in x! 24 249 2499 24999 249998

__Question 47 - magic.squares:__

Are there large squares, containing only consecutive integers, all of whose
rows, columns and diagonals have the same sum? How about cubes?
Show Answer

These are called magic squares. A magic square of order n (integers from 1 to n*n) has only one possible sum: (n*n+1)*n/2.Odd and even order squares must be constructed by different approaches. For odd orders, the most common algorithm is a recursive scheme devised by de la Loubere about 300 years ago. For even orders, one procedure is the Devedec algorithm, which treats even orders not divisible by 4 slightly differently from those which are divisible by 4 (doubly even).

For squares with odd-length sides, the following algorithm builds a magic square:

Put 1 in the middle box in the upper row. From then on, if it's possible to put the next number one box diagonally up and to the right (wrapping around if the edge of the grid is reached), do so, otherwise, put it directly below the last one.

17 24 1 8 15

23 5 7 14 16

4 6 13 20 22

10 12 19 21 3

11 18 25 2 9

...or even47 58 69 80 1 12 23 34 45

57 68 79 9 11 22 33 44 46

67 78 8 10 21 32 43 54 56

77 7 18 20 31 42 53 55 66

6 17 19 30 41 52 63 65 76

16 27 29 40 51 62 64 75 5

26 28 39 50 61 72 74 4 15

36 38 49 60 71 73 3 14 25

37 48 59 70 81 2 13 24 35See archive entry knight.tour for magic squares that are knight's tours.

To get a 4x4 square, write the numbers in order across each row, filling the square...

1 2 3 4

5 6 7 8

9 10 11 12

13 14 15 16then use the following pattern as a mask:

. X X .

X . . X

X . . X

. X X .Everywhere there is an X, complement the number (subtract it from n*n+1). For the 4x4 you get:

1 15 14 4

12 6 7 9

8 10 11 5

13 3 2 16For n even (n>4):

Make an initial magic square by writing an n/2 magic square four times (the same one each time). Now, although this square adds up right we have the numbers 1 to n*n/4 written four times each. To fix this, simply add to it n*n/4 times one of the following magic squares:

if n/2 is odd (example: n/2=7),

3 3 3 0 0 0 0 2 2 2 2 2 1 1 (there are int(n/4) 3s, int(n/4-1) 1s on each

3 3 3 0 0 0 0 2 2 2 2 2 1 1 row)

3 3 3 0 0 0 0 2 2 2 2 2 1 1

0 3 3 3 0 0 0 2 2 2 2 2 1 1 (this is row int(n/4)+1. It starts with just

3 3 3 0 0 0 0 2 2 2 2 2 1 1 the one 0)

3 3 3 0 0 0 0 2 2 2 2 2 1 1

3 3 3 0 0 0 0 2 2 2 2 2 1 1

0 0 0 3 3 3 3 1 1 1 1 1 2 2 (the lower half is the same as the upper half

0 0 0 3 3 3 3 1 1 1 1 1 2 2 with 3<->0 and 1<->2 swapped. This guarantees

0 0 0 3 3 3 3 1 1 1 1 1 2 2 that each number 1-n*n will appear in the

3 0 0 0 3 3 3 1 1 1 1 1 2 2 completed square)

0 0 0 3 3 3 3 1 1 1 1 1 2 2

0 0 0 3 3 3 3 1 1 1 1 1 2 2

0 0 0 3 3 3 3 1 1 1 1 1 2 2if n/2 is even (example: n/2=4), (there are n/4 of each number on each row)

0 0 3 3 2 2 1 1

0 0 3 3 2 2 1 1

0 0 3 3 2 2 1 1

0 0 3 3 2 2 1 1

3 3 0 0 1 1 2 2

3 3 0 0 1 1 2 2

3 3 0 0 1 1 2 2

3 3 0 0 1 1 2 2References:

"Magic Squares and Cubes"

W.S. Andrews

The Open Court Publishing Co.

Chicago, 1908

"Mathematical Recreations"

M. Kraitchik

Dover

New York, 1953

__Question 48 - pell:__

Find integer solutions to x^2 - 92y^2 = 1.
Show Answer

x=1 y=0

x=1151 y=120

x=2649601 y=276240

etc.Each successive solution is about 2300 times the previous solution; they are every 8th partial fraction (x=numerator, y=denominator) of the continued fraction for sqrt(92) =

[9, 1,1,2,4,2,1,1,18, 1,1,2,4,2,1,1,18, 1,1,2,4,2,1,1,18, ...]Once you have the smallest positive solution (x1,y1) you don't need to "search" for the rest. You can obtain the nth positive solution (xn,yn) by the formula

(x1 + y1 sqrt(92))^n = xn + yn sqrt(92).

See Niven & Zuckerman's An Introduction to the Theory of Numbers. Look in the index under Pell's equation.

__Question 49 - subset:__

Prove that all sets of n integers contain a subset whose sum is divisible by n.
Show Answer

Consider the set of remainders of the partial sums a(1) + ... + a(i). Since there are n such sums, either one has remainder zero (and we're thru) or 2 coincide, say the i'th and j'th. In this case, a(i+1) + ... + a(j) is divisible by n. (note this is a stronger result since the subsequence constructed is of *adjacent* terms.) Consider a(1) (mod n), a(1)+a(2) (mod n), ..., a(1)+...+a(n) (mod n). Either at some point we have a(1)+...+a(m) = 0 (mod n) or else by the pigeonhole principle some value (mod n) will have been duplicated. We win either way.

__Question 50 - sum.of.cubes:__

Find two fractions whose cubes total 6.
Show Answer

Restated:

Find X, Y, minimum Z (all positive integers) where

(X/Z)^3 + (Y/Z)^3 = 6Again, a generalized solution would be nice.

You are asking for the smallest z s.t. x^3 + y^3 = 6*z^3 and x,y,z in Z+. In general, questions like these are extremely difficult; if you're interested take a look at books covering Diophantine equations (especially Baker's work on effective methods of computing solutions).

Dudeney mentions this problem in connection with #20 in _The Canterbury Puzzles_; the smallest answer is (17/21)^3 + (37/21)^3 = 6.

For the interest of the readers of this group I'll quote:

"Given a known case for the expression of a number as the sum or difference of two cubes, we can, by formula, derive from it an infinite number of other cases alternately positive and negative. Thus Fermat, starting from the known case 1^3 + 2^3 = 9 (which we will call a fundamental case), first obtained a negative solution in bigger figures, and from this his positive solution in bigger figures still. But there is an infinite number of fundamentals, and I found by trial a negative fundamental solution in smaller figures than his derived negative solution, from which I obtained the result shown above. That is the simple explanation."

In the above paragraph Dudeney is explaining how he derived (*by hand*) that (415280564497/348671682660)^3 + (676702467503/348671682660)^3 = 9.

He continues:

"We can say of any number up to 100 whether it is possible or not to express it as the sum of two cubes, except 66. Students should read the Introduction to Lucas's _Theorie des Nombres_, p. xxx."

"Some years ago I published a solution for the case 6 = (17/21)^3 + (37/21)^3, of which Legendre gave at some length a 'proof' of impossibility; but I have since found that Lucas anticipated me in a communication to Sylvester."

__Question 51 - sums.of.powers:__

Partition 1,2,3,...,16 into two equal sets, such that the sums of the
numbers in each set are equal, as are the sums of their squares and cubes.
Show Answer

The solution is

A = {1,4,6,7,10,11,13,16}

B = {2,3,5,8,9,12,14,15}Let X+k be a set formed by adding k to all elements of X. Then A+k and B+k have the property of satisfying i,ii and iii. That means, any 16 numbers in A.P can be partioned in such a way to satisfy i,ii and iii.

How do we arrive at the above solution without using a computer?

By the preceding discussion,

A1 = A-1 = {0,3,5,6,9,10,12,15}

B1 = B-1 = {1,2,4,7,8,11,13,14}have the property of satisfying i,ii and iii. Notice that all numbers of A1 have even number of 1's in binary and all numbers of B1 have odd number of 1's in binary. Essentially the above partition is a partition based on parity.

Observation:

Partition of (n+1) bit numbers based on parity into two groups A and B (each consisting of 2^n elements) satisfies

sum of kth powers of elements of A = sum of kth powers of elements of B

for k=1,2,...,n. (The above puzzle is a special case n=3).

The above observation also holds for any radix. In radix r we will have r groups.

Infact,

Any r^(n+1) terms in A.P can be partitioned into r groups (each of r^n elements) such that sum of kth powers of all r groups is same (k=1,2,...,n)

Finding such groups with minimal number of elements (less than r^n) appears to be more difficult!

ALL THIS APPEARS TO BE INTERESTING. IS IT A CONSEQUENCE OF A SIMPLE THEOREM OF NUMBER THEORY? HOW DO I PROVE THIS?

Thanks in advance for any clues,

-- ramana@svl.cdc.com (Mr. Ramana (Indian analyst))

__Question 52 - tests.for.divisibility/eleven:__

What is the test to see if a number is divisible by eleven?
Show Answer

If the alternating sum of the digits is divisible by eleven, so is the number.For example, 1639 leads to 9 - 3 + 6 - 1 = 11, so 1639 is divisible by 11.

Proof:

Every integer n can be expressed as

n = a1*(10^k) + a2*(10^k-1)+ .....+ a_k+1

where a1, a2, a3, ...a_k+1 are integers between 0 and 9.

10 is congruent to -1 mod(11).

Thus if (-1^k)*a1 + (-1^k-1)*a2 + ...+ (a_k+1) is congruent to 0mod(11) then

n is divisible by 11.

__Question 53 - tests.for.divisibility/nine:__

What is the test to see if a number is divisible by nine?
Show Answer

If the sum of the digits is divisible by nine, so is the number.

Proof:

Every integer n can be expressed as

n = a1*(10^k) + a2*(10^k-1)+ .....+ a_k+1

where a1, a2, a3, ...a_k+1 are integers between 0 and 9.

Note that 10 is congruent to 1 (mod 9). Thus 10^k is congruent to 1 (mod 9) for

every k >= 0.

Thus n is congruent to (a1+a2+a3+....+a_k+1) mod(9).

Hence (a1+a2+...+a_k+1) is divisible by 9 iff n is divisible by 9.

__Question 54 - tests.for.divisibility/seven:__

What is the test to see if a number is divisible by seven?
Show Answer

Take the last digit (n mod 10) and double it.

Take the rest of the digits (n div 10) and subtract the doubled last digit from it.

The resulting number is divisible by 7 iff the original number is divisible by 7.

Example: Take 53445

Subtract (53445 mod 10) * 2 from (53445 div 10)

- 5 * 2 + 5344

= 5334

533 - 2 * 4 = 525

52 - 5 * 2 = 42 which is divisible by 7

so 53445 is divisible by 7.

__Question 55 - tests.for.divisibility/three__

What is the test to see if a number is divisible by three?
Show Answer

A number is divisible by three iff the sum of its digits is divisible by three.

First, prove 10^N = 1 mod 3 for all integers N >= 0. 1 = 1 mod 3. 10 = 1 mod 3. 10^N = 10^(N-1) * 10 = 10^(N-1) mod 3.

QED by induction.

Now let D[0] be the units digit of N, D[1] the tens digit, etc.

Now N = Summation From k=0 to k=inf of D[k]*10^k.

Therefore N mod 3 = Summation from k=0 to k=inf of D[k] mod 3. QED