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`        Prove 11^n+2+12^2n+1 is divisible by 133 using binomial theorem`
one year ago

Arun
23044 Points
```							Dear student PART 1: First, prove it for the base case of n = 1: 11^(1+2) + 12^(2+1) = 11^3 + 12^3 = 1331 + 1728 = 133(10) + 1 + 1728 = 133(10) + 1729 = 133(10) + 133(13) = 133(26) PART 2: Assume it is true for a natural number k. Prove it is true for the number k+1: True for k: 11^(k+2) + 12^(2k+1) = 133(m), where m is an integer. For k+1: 11^(k+1+2) + 12^(2(k+1)+1) = 11^(k+2) * 11 + 12^(2k + 2 + 1) = 11^(k+2) * 11 + 12^(2k + 1) * 12² = 11 * 11^(k+2) + (11 + 133) * 12^(2k+1) = 11( 11^(k+2) + 12^(2k+1) ) + 133 * 12^(2k+1) = 11( 133m ) + 133 * 12^(2k+1) = 133 ( 11m + 12^(2k + 1)) The stuff in the parentheses will be an integer, so the result is a multiple of 133. Therefore by induction a number of that form is divisible by 133.    RegardsArun (askIITians forum expert)
```
one year ago
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