A Binomial Coefficient Question

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

#BinomialCoefficients #TorqueGroup

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 $\boxed{16}$

Pi Han Goh - 7 years, 5 months ago

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

Josh Rowley - 7 years, 5 months ago

Fixed......... THANKS

Pi Han Goh - 7 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$