# Difference between revisions of "2020 CIME I Problems/Problem 7"

## Problem 7

For every positive integer $n$, define $$f(n)=\frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1)}.$$ Suppose that the sum $f(1)+f(2)+\cdots+f(2020)$ can be expressed as $\frac{p}{q}$ for relatively prime integers $p$ and $q$. Find the remainder when $p$ is divided by $1000$.

## Solution

Let $S(n) = f(1)+f(2)+\cdots+f(n)$. We claim that $$S(n) = \frac{1}{2} - \frac{1}{2 \cdot (2n+1)!!}.$$ We show this using induction. Suppose this is true for some $n = k$. Then, it must be true for $k+1$. The base case when $k = 1$ is trivial. Then, we have that \begin{align*} S(k+1) &= f(k) + S(k) \\ {} &= \frac{n}{(2n+1)!!} + \frac{1}{2} - \frac{1}{2 \cdot (2n+1)!!} \\ {} &= \frac{1}{2} - \frac{1}{2 \cdot (2n+3)!!}. \end{align*} Hence, this completes the induction therefore proving the claim. So, the numerator $p$ is $\frac{1}{2}(4041!! - 1)$. We proceed using Euler's Theorem combined with Chinese Remainder Theorem. It is obvious that $4041!! \equiv 0 \pmod{125}$ so $p \equiv 62\pmod{125}$. Also, instead of considering $p$ modulo $8$, we consider it modulo $16$. Then, we get $$4041!! \equiv 3^{253} \cdot 5^{253} \cdot 7^{253} \cdot 9^{252} \cdot 11^{252} \cdot 13^{252} \cdot 15^{252} \equiv 3 \cdot 5 \cdot 7 \cdot 9 \equiv 1\pmod{16},$$ by Euler's Totient Theorem as $\phi(16) = 8$. This implies that $\frac{1}{2}(4041!! - 1) \equiv 0\pmod{8}$, so $p \equiv 0\pmod{8}$. Solving the system of congruences, we get $p \equiv \boxed{312}\pmod{1000}$. ~rocketsri (based off of official solution)

## See also

 2020 CIME I (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All CIME Problems and Solutions

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