Even the answer is a palindrome

Number Theory Level 5

Find

( ∑ k = 0 1221 ( 1 1 k ( 1221 ) 1 1 k k ) ) ( m o d 1331 ) \displaystyle \bigg( \sum_{k=0}^{1221} {11^{k}(1221) \choose 11^{k}k} \bigg) \pmod{1331}


The answer is 717.

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Jake Lai by Jake Lai , 21, Singapore

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1 solution

Jake Lai
Dec 14, 2014

Use Wolstenholme's theorem k k times on each term.

The sum reduces to 2 1221 ≡ 2 11 ≡ 717 ( m o d 1331 ) 2^{1221} \equiv 2^{11} \equiv 717 \pmod{1331} by Euler's theorem.

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