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Frequently Asked Questions (FAQS);faqs.454 State Symbol New Symbol New Symbol New Accept? State State 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 Symbol New Symbol New Symbol New Accept? State State 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 Symbol New Symbol New Symbol New Accept? State State 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 ==> arithmetic/digits/power.two.p <== Prove that for any 9-digit number (base 10) there is an integral power of 2 whose first 9 digits are that number. ==> arithmetic/digits/power.two.s <== 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. ==> arithmetic/digits/prime/101.p <== How many primes are in the sequence 101, 10101, 1010101, ...? ==> arithmetic/digits/prime/101.s <== 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 thus 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. ==> arithmetic/digits/prime/all.prefix.p <== What is the longest prime whose every proper prefix is a prime? ==> arithmetic/digits/prime/all.prefix.s <== 23399339, 29399999, 37337999, 59393339, 73939133 ==> arithmetic/digits/prime/change.one.p <== What is the smallest number that cannot be made prime by changing a single digit? Are there infinitely many such numbers? ==> arithmetic/digits/prime/change.one.s <== 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. ==> arithmetic/digits/prime/prefix.one.p <== 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? ==> arithmetic/digits/prime/prefix.one.s <== 149 ==> arithmetic/digits/reverse.p <== Is there an integer that has its digits reversed after dividing it by 2? ==> arithmetic/digits/reverse.s <== 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. ==> arithmetic/digits/rotate.p <== Find integers where multiplying them by single digits rotates their digits. ==> arithmetic/digits/rotate.s <== 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 10112359550561797752808988764044943820224719 In 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 - 1 In 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)<R, and the set of all solutions is the set of multiples of N0(R,B). N0(R,B) is the length of the repeating part of the fraction 1/R in base 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. ==> arithmetic/digits/sesqui.p <== Find the least number where moving the first digit to the end multiplies by 1.5. Xref: bloom-picayune.mit.edu rec.puzzles:18138 news.answers:3070 Newsgroups: rec.puzzles,news.answers Path: bloom-picayune.mit.edu!snorkelwacker.mit.edu!spool.mu.edu!uunet!questrel!chris From: uunet!questrel!chris (Chris Cole) Subject: rec.puzzles FAQ, part 3 of 15 Message-ID: <puzzles-faq-3_717034101@questrel.com> Followup-To: rec.puzzles Summary: This posting contains a list of Frequently Asked Questions (and their answers). It should be read by anyone who wishes to post to the rec.puzzles newsgroup. Sender: chris@questrel.com (Chris Cole) Reply-To: uunet!questrel!faql-comment Organization: Questrel, Inc. References: <puzzles-faq-1_717034101@questrel.com> Date: Mon, 21 Sep 1992 00:08:46 GMT Approved: news-answers-request@MIT.Edu Expires: Sat, 3 Apr 1993 00:08:21 GMT Lines: 1353 Archive-name: puzzles-faq/part03 Last-modified: 1992/09/20 Version: 3 ==> arithmetic/digits/sesqui.s <== 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. ==> arithmetic/digits/squares/leading.7.to.8.p <== What is the smallest square with leading digit 7 which remains a square when leading 7 is replaced by an 8? ==> arithmetic/digits/squares/leading.7.to.8.s <== x=2996282391593370361328125 y=2824483699753370361328125 x^2=8977708170172487211329625006796419620513916015625 y^2=7977708170172487211329625006796419620513916015625 ==> arithmetic/digits/squares/length.22.p <== 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. ==> arithmetic/digits/squares/length.22.s <== No, the number of digits of A^2 must be of the form 3n or 3n-1. ==> arithmetic/digits/squares/length.9.p <== Is it possible to make a number and its square, using the digits from 1 through 9 exactly once? ==> arithmetic/digits/squares/length.9.s <== 567 and 854. ==> arithmetic/digits/squares/three.digits.p <== What squares consist entirely of three digits (e.g., 1, 4, and 9)? ==> arithmetic/digits/squares/three.digits.s <== The full set of solutions up to 10**12 is 1 -> 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.uk Hey guys, what about 648070211589107021 ^ 2 = 419994999149149944149149944191494441 This 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!guy Here 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 = 1 while(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 ==> arithmetic/digits/squares/twin.p <== Let a twin be a number formed by writing the same number twice, for instance, 81708170 or 132132. What is the smallest square twin? ==> arithmetic/digits/squares/twin.s <== 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 a < 10^n. If ak is a perfect square, k must have some repeated factor, since a<k. Searching the possible values of k for one with a repeated factor eventually turns up the number 1 + 10^11 = 11^2 * 826446281. So, we set a=826446281 and ak = 9090909091^2 = 82644628100826446281, but this needs leading zeros to fit the pattern. So, we multiply by a suitable small square (in this case 16) to get the above answer. ==> arithmetic/digits/sum.of.digits.p <== Find sod ( sod ( sod (4444 ^ 4444 ) ) ). ==> arithmetic/digits/sum.of.digits.s <== let X = 4444^4444 sod(X) <= 9 * (# of digits) < 145900 sod(sod(X)) <= sod(99999) = 45 sod(sod(sod(X))) <= sod(39) = 12 but sod(sod(sod(X))) = 7 (mod 9) thus sod(sod(sod(X))) = 7 ==> arithmetic/digits/zeros/factorial.p <== How many zeros are in the decimal expansion of n!? ==> arithmetic/digits/zeros/factorial.s <== 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 ==> arithmetic/digits/zeros/lsd.factorial.p <== What is the least significant non-zero digit in the decimal expansion of n!? ==> arithmetic/digits/zeros/lsd.factorial.s <== 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<stdio.h> #include<assert.h> 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]; } ==> arithmetic/digits/zeros/million.p <== How many zeros occur in the numbers from 1 to 1,000,000? ==> arithmetic/digits/zeros/million.s <== 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, so p(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. ==> arithmetic/magic.squares.p <== Are there large squares, containing only consecutive integers, all of whose rows, columns and diagonals have the same sum? How about cubes? ==> arithmetic/magic.squares.s <== Here is an 8x8 example: 01 56 48 25 33 24 16 57 63 10 18 39 31 42 50 07 62 11 19 38 30 43 51 06 04 53 45 28 36 21 13 60 05 52 44 29 37 20 12 61 59 14 22 35 27 46 54 03 58 15 23 34 26 47 55 02 08 49 41 32 40 17 09 64 References: "Magic Squares and Cubes" W.S. Andrews The Open Court Publishing Co. Chicago, 1908 "Mathematical Recreations" M. Kraitchik Dover New York, 1953 ==> arithmetic/pell.p <== Find integer solutions to x^2 - 92y^2 = 1. ==> arithmetic/pell.s <== 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. ==> arithmetic/prime/arithmetic.progression.p <== Is there an arithmetic progression of 20 or more primes? ==> arithmetic/prime/arithmetic.progression.s <== 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. ==> arithmetic/prime/consecutive.composites.p <== Are there 10,000 consecutive non-prime numbers? ==> arithmetic/prime/consecutive.composites.s <== 9973!+2 through 9973!+10006 are composite. ==> arithmetic/sequence.p <== Prove that all sets of n integers contain a subset whose sum is divisible by n. ==> arithmetic/sequence.s <== 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. ==> arithmetic/sum.of.cubes.p <== Find two fractions whose cubes total 6. ==> arithmetic/sum.of.cubes.s <== Restated: Find X, Y, minimum Z (all positive integers) where (X/Z)^3 + (Y/Z)^3 = 6 Again, 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." ==> arithmetic/tests.for.divisibility/eleven.p <== What is the test to see if a number is divisible by eleven? ==> arithmetic/tests.for.divisibility/eleven.s <== 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. ==> arithmetic/tests.for.divisibility/nine.p <== What is the test to see if a number is divisible by nine? ==> arithmetic/tests.for.divisibility/nine.s <== 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.