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in Mathematics by (155 points)
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Prove that 11n+2 + 122n+1 is divisible by 133? 

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1 Answer

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11n+2 + 122n+1 

for n = 1 we get

112 + 123 = 1331 + 1728 = 3058

3059/133 = 23

hence it is true for n = 1

let us assume that this expression is true for n = k

11k+2 + 122k + 1  is divisible by 133 for n = k + 1

11k + 3 + 122k + 3 = (11x 1331) + (122k x 1728) = (11k x 1331) + (11 + 1)2k x 1728)

which is divisible by 1331 + 1728 = 3059 which is divisible by 133.

Hence our assumption is correct and 11n+2 + 122n+1 is divisible by 133.

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