Mathematics.Mathematical Jokes and Puzzles

Mathematics Forum

Author	Message / Information
Soroban Quote \| Reply \| This message was updated on 8/31/2002 10:41:35 PM by Soroban	Challenger #1 posted on: 8/31/2002 10:38:47 PM Prove that, for any positive integer n, 2903^n - 803^n - 464^n + 261^n is divisible by 1897.
Soroban Quote \| Reply \| This message was updated on 9/10/2002 2:14:17 PM by Soroban	Challenger #1 - Solution replied on: 9/10/2002 1:54:23 PM No takers! ~ Too hard? ~ Too easy? ~ Not interested? Let C = 2903^n - 803^n - 464^n + 261^n The solution is based on the identity: (a^n - b^n)/(a - b) = = a^(n-1) + a^(n-2)b + a^(n-3)b^2 +... + ab^(n-2) + b^(n-1) That is, the difference of the nth power of two integers is divisible by the difference of the two integers. Consider: C = (2903^n - 803^n) - (464^n - 261^n) ~ The first group is divisible by 2903 - 803 = 2100 = 7300. ~ The second group is divisible by 464-261 = 203 = 729. Hence, C is divisible by 7. Consider: C = (2903^n - 464^n) - (803^n - 261^n) ~ The first group is divisible by 2903 - 464 = 2439 = 9271. ~ The second group is divisible by 803 - 261 = 542 = 2*271. Hence, C is divisible by 271. Since C is divisible by 7 and by 271, it is divisible by 7271 = 1897*.
LinkBot	Gamers Wanted is looking for people to write game reviews and post news, if your interested please visit Gamers Wanted About Us Page