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3. Problem: Show that 49" +16n -1 is divisible by 64 for all positive integers n. |
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Answer» OK, I'm assuming anyone reading this will be familiar with the concept a mod b... If you're not, a mod b is the remainder you'd get dividing a by b. I'm going to use = for "congruent to" here... What this boils down to proving is that: 49^n + 16n = 1 mod 64 Let's start with the easy stuff... Power of 49 49^n mod 640 11 492 333 174 1 WELL! Now, THAT simplifies things a lot, doesn't it? The index of 49 mod 64 is merely 4... Thus, for all n, let n = a mod 4. Then, we can say that 49^n = 49^a mod 64. OK, now all we have to do is get this 16 n thing in there... When you do that, you get: 49^n mod 64 49^ n + 16n mod 641 149 49 + 16 = 6533 33 + 32 = 6517 17 + 48 = 651 1 + 64 = 65 In all of these cases, subtract 1 and you get 64... |
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