Ed, "Find positive integers A, B, C, D, and E all less than 100 so that A^2 + B^2 + C^2 = D^2 + E^2 and A^3 + B^3 + C^3 = D^3 + E^3" Answer: For integers from 1-99, there is one unique solution set not counting swapping A, B, C around, or swapping D and E: For A,B,C, the numbers are 21, 26, 64. For D and E, the numbers are 37 and 62. Counting all combinations of these due to addition reordering, there are 12 solution combinations. Further analysis: Counting all possible combinations, there are 647913 solutions to A^2 + B^2 + C^2 = D^2 + E^2, and 11550 solutions to A^3 + B^3 + C^3 = D^3 + E^3. If we change the range from 1-99 to 1-199, there are three solution sets (or 36 total solution combinations counting addition reordering): 1. (A,B,C)=42,52,128; (C,D)=74,124 2. (A,B,C)=63,78,192; (C,D)= 111,186 3. (A,B,C)=21,26,64; (C,D)=37,62 Thanks for the diversion on a slow work day! Happy Holidays, J.B. Gill ----------------------------------------------------------------------- --- A, B, C, D, and E all less than 100 so that A2 + B2 + C2 = D2 + E2 and A3 + B3 + C3 = D3 + E3 are: A=21, B=26, C=64 or a permutation of these. and D=37, E=62 or D=62 and E=37. Regarding ending of The square of 40081787109376 which starts like 16065496578813... is in the same digit as the origional number i.e. 40081787109376 , so the complete square is 1606549657881340081787109376, it is a automorphic number. Shyam Sunder Gupta www.shyamsundergupta.com guptass@rediffmail.com ---------------------------------------------------- Hi Ed, I was unable to find anything else than (21,26,64) and (37,62) as solutions. KR, Jean-François Halleux ----------------------------------------------------------------------- - Erich Friedman: Find positive integers A, B, C, D, and E all less than 100 so that A2 + B2 + C2 = D2 + E2 and A3 + B3 + C3 = D3 + E3. Solution: a=21 b=26 c=64 d=37 e=62 Dave Gao ------------------------------------------------------------ The only solution which I have found to the problem, with a hurried program, is: 21^2 +26^2 +64^2 = 37^2 +62^2 = 5213 21^3 +26^3 +64^3 = 37^3 +62^3 = 288981 Happy New Year Jordi Domènech -------------------------------------------------------------- My answer to the puzzle A^2+B^2+C^2=D^2+E^2 and A^3+B^3+C^3=D^3+E^3 is as follows: A=21 B=26 C=64 D=37 E=62 I really enjoy your puzzle page! Bill Estabrook ------------------------------------------------------------- sorry, had a little floatingpoint-inaccuracy in my proggy.. 21² + 26² + 64² = 37² + 62² 21³ + 26³ + 64³ = 37³ + 62³ Franz. p.s.: can this be solved algebraic ? --------------------------------------------------------------- 21^2 + 26^2 + 64^2 = 37^2 + 62^2 21^3 + 26^3 + 64^3 = 37^3 + 62^3 Who invented that this is possible? Must be a genious! Juha Saukkola --------------------------------------------------------------- "Find positive integers A, B, C, D, and E all less than 100 so that A^2 + B^2 + C^2 = D^2 + E^2 and A^3 + B^3 + C^3 = D^3 + E^3." I think I have found the only answer. A=21 B=26 C=64 D=37 E=62 Since A, B, & C are interchangeable, any of the values for those variables can be switched (e.g. (A,B,C) = (26,21,64) is also acceptable. The same is true for D & E. I used Microsoft Excel to solve this one, and it took me hours. Alan Lemm ---------------------------------------------------------- Find positive integers A, B, C, D, and E all less than 100 so that A2 + B2 + C2 = D2 + E2 and A3 + B3 + C3 = D3 + E3. Answer: A = 21, B = 26, C = 64, D = 37, E = 62 Daniel Lidström ------------------------------------------------------------- Solution: 21, 26, 64, 37, 62 are values for A, B, C, D, and E that satisfy the two conditions: A^2 + B^2 + C^2 = D^2 + E^2 and A^3 + B^3 + C^3 = D^3 + E^3 I used a brute force approach with Excel, first creating a million item spread sheet containing all of the possible "D^3+E^3-C^3," then searching it for any results equivalent to one of the 10,000 possible "A^3 + B^3" sums, and lastly testing the coordinates that yielded matches in the squares equation. I'm curious to know of more insightful mathier strategies, and what patterns appear among the two sets of solutions to help find their intersection. So back to the spreadsheets to see what I can see. - Matt Sheppeck While I was marvelling at the speed with which my Christmas laptop can perform 100,000,000 comparisons, Tom Hanks was talking of amazing room-size NASA computers performing thousands of calculations in Apollo 13. ------------------------------------------------------------------- 200:3 -> 400:0:9 -> 1600:8:1 -> 2:560065 -> 3:1:36:7:2:8:0:4:22:9 -> 2004 200:3 -> 4000:9 -> 1600:0:0:81 -> 2:560065 -> 3:1:36:7:2:8:0:4:22:9 -> 2004 200:3 -> 4000:9 -> 16000:0:81 -> 2:5:60065:6:1 -> 3:6:0:7:8:0:42:9:1 -> 2004 200:3 -> 4000:9 -> 1:600008:1 -> 3600:0:9:600066 -> 3:6:0:0:9:21:6:4:4:37 -> 2004 200:3 -> 4000:9 -> 1600008:1 -> 25:60:0:25:60006::5 -> 36:0:0:7:24:9:1:1 -> 2004 200:3 -> 4000:9 -> 1:6000081 -> 3:6000:97:2006:5:6:2 -> 40:0:3:3:5:19 -> 2004 200:3 -> 4000:9 -> 160000:81 -> 2:5:6000065:6:1 -> 3:6:0:0:0:7:8:0:0:0:42:9:1 -> 2004 And I'm sure there are a lot more =) Yogy Namara --------------------------------------------------------------------- Hi Ed The Erich Friedman problem A^2 + B^2 + C^2 = D^2 + E^2 and A^3 + B^3 + C^3=D^3 + E^3 Solution A=21 B=26 C=64 D=37 E=62 Regards Paul Cleary.