Archivename: puzzles/archive/arithmetic/part2
Lastmodified: 17 Aug 1993 Version: 4 See reader questions & answers on this topic!  Help others by sharing your knowledge ==> 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 depthfirst 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 ndigit solutions (i.e. 2ndigit squares) using this lastsignificantdigitfirst 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 mostsignificantdigitfirst 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@nsfnetrelay.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 149property; checking took a bit more than an hour of CPU time. As somebody suggested, we used a combined mostsignificant/leastsignificant digits attack. First we make a table of pdigit 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 singleended algorithm, this lets us check 50% more digits in the same amount of time (ignoring polynomial factors). Of course, the space cost of the combinedends 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@csnetrelay (412) 2683056 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 10^(n1) <= 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/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^(n1) through 10^n  1, there are 9 * 10^(n1) numbers of n digits each, so 9(n1)10^(n1) nonleading digits, of which one tenth, or 9(n1)10^(n2), are zeroes. When we change the range to 10^(n1) + 1 through 10^n, we remove 10^(n1) and put in 10^n, gaining one zero, so p(n) = p(n1) + 9(n1)10^(n2) + 1 with p(1)=1. Solving the recurrence yields the closed form p(n) = n(10^(n1)+1)  (10^n1)/9. For n=6, there are 488,895 zeroes, 600,001 ones, and 600,000 of all other digits. ==> arithmetic/digits/zeros/trailing.p <== How many trailing zeros are in the decimal expansion of n!? ==> arithmetic/digits/zeros/trailing.s <== The general answer to the question "what power of p divides x!" where p is prime is (xd)/(p1) 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 = (102)/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/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 <== 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 oddlength 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 even 47 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 35 See 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 16 then 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 16 For 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/41) 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 1n*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 2 if n/2 is even (example: n/2=4), 0 0 3 3 2 2 1 1 (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 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 2 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/subset.p <== Prove that all sets of n integers contain a subset whose sum is divisible by n. ==> arithmetic/subset.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/sums.of.powers.p <== 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. ==> arithmetic/sums.of.powers.s <== 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 = A1 = {0,3,5,6,9,10,12,15} B1 = B1 = {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)) ==> 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^k1)+ .....+ 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^k1)*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^k1)+ .....+ 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. ==> arithmetic/tests.for.divisibility/seven.p <== What is the test to see if a number is divisible by seven? ==> arithmetic/tests.for.divisibility/seven.s <== 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. ==> arithmetic/tests.for.divisibility/three.p <== What is the test to see if a number is divisible by three? ==> arithmetic/tests.for.divisibility/three.s <== 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^(N1) * 10 = 10^(N1) 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 User Contributions:Comment about this article, ask questions, or add new information about this topic:Part1  Part2 [ Usenet FAQs  Web FAQs  Documents  RFC Index ] Send corrections/additions to the FAQ Maintainer: archivecomment@questrel.com
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