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# 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 $$30$$.

Note by Anqi Li
3 years, 8 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 $$\boxed{16}$$ · 3 years, 8 months ago

Do you mean $$\displaystyle\sum_{k=0}^{n} \dbinom {2n}{2k} = 2^{2n-1}$$? · 3 years, 8 months ago