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Find the remainder when \( {1234 \choose 2} + {1234 \choose 6} + {1234 \choose 10} \ldots + {1234 \choose 1230} + {1234 \choose 1234} \) is divided by \( 30 \).

Note by Anqi Li 3 years, 11 months ago

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Let \(N \) denote that expression, because \( {n \choose k } = {n \choose n-k} \)

\(N = {1234 \choose 1232 } + {1234 \choose 1228 } + {1234 \choose 1224 } + \ldots + {1234 \choose 4 } + {1234 \choose 0 } \)

Add the "original" \(N\) and the above equation

\( 2N = {1234 \choose 0 } + {1234 \choose 2 } + {1234 \choose 4 } + {1234 \choose 6 } + \ldots + {1234 \choose 1230 } + {1234 \choose 1232 } + {1234 \choose 1234 } \)

Because \( \displaystyle \sum_{k=0}^n {2n \choose 2k} = 2^{2n-1} \)

\( 2N = 2^{1233} \)

\( N = 2^{1232} \)

\( N \pmod {2} \equiv 0 \)

\( N \pmod {3} \equiv (-1)^{1232} \equiv 1 \)

\( N \pmod {5} \equiv 2^{ 1232 \space \bmod {4} } \equiv 2^0 \equiv 1 \), by Fermat's Little Theorem

Because \(2,3,5\) are pairwise coprime, we can find \(N \) modulo \(30\) by Chinese Remainder Theorem, the answer is \(\boxed{16} \)

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Do you mean \( \displaystyle\sum_{k=0}^{n} \dbinom {2n}{2k} = 2^{2n-1} \)?

Fixed......... THANKS

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`\sin \theta`

`\boxed{123}`

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TopNewestLet \(N \) denote that expression, because \( {n \choose k } = {n \choose n-k} \)

\(N = {1234 \choose 1232 } + {1234 \choose 1228 } + {1234 \choose 1224 } + \ldots + {1234 \choose 4 } + {1234 \choose 0 } \)

Add the "original" \(N\) and the above equation

\( 2N = {1234 \choose 0 } + {1234 \choose 2 } + {1234 \choose 4 } + {1234 \choose 6 } + \ldots + {1234 \choose 1230 } + {1234 \choose 1232 } + {1234 \choose 1234 } \)

Because \( \displaystyle \sum_{k=0}^n {2n \choose 2k} = 2^{2n-1} \)

\( 2N = 2^{1233} \)

\( N = 2^{1232} \)

\( N \pmod {2} \equiv 0 \)

\( N \pmod {3} \equiv (-1)^{1232} \equiv 1 \)

\( N \pmod {5} \equiv 2^{ 1232 \space \bmod {4} } \equiv 2^0 \equiv 1 \), by Fermat's Little Theorem

Because \(2,3,5\) are pairwise coprime, we can find \(N \) modulo \(30\) by Chinese Remainder Theorem, the answer is \(\boxed{16} \)

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Do you mean \( \displaystyle\sum_{k=0}^{n} \dbinom {2n}{2k} = 2^{2n-1} \)?

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Fixed......... THANKS

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