Massive Binomial Product

Number Theory Level pending

Find the smallest positive integer $n$ for which the number $A_n = \prod_{k=1}^n \binom{k^2}{k} = \binom{1}{1} \binom{4}{2} \cdots \binom{n^2}{n}$ ends in the digit $0$ when written in base ten.

The answer is 4.

1 solution

Sharky Kesa
Aug 20, 2017

Note that $\dbinom{4}{2}$ has a factor of $2$ and $\dbinom{16}{4}$ has a factor of $5$ . Thus, $A_4$ satisfies being divisible by $2$ and $5$ , so $A_4$ is divisible by 10, so it ends in 0. Thus, $n=4$ is the smallest number.

Massive Binomial Product

The answer is 4.

1 solution

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