#17 of June 2018 Grade 11 - Natural Sciences CSAT (Korean SAT) Mock test

Calculus Level 5

A sequence $\{a_n\}$ with $a_1=1,~a_2=-1,~a_3=32$ satisfies

$n(n-2)a_{n+1}=\sum_{i=1}^{n}a_i$

for all positive integers $n.$

Let $S(n)$ denote the number of positive integers $k\le n$ such that $\dfrac{1}{a_k}$ is an integer.

Given $S(10^{20})=p\times 10^{q}.$ where $p,~q$ are positive integers, with $p$ not a multiple of $10,$ find $p+q.$

This problem has been modified by me, and is a part of <Grade 11 CSAT Mock test> series .

The answer is 3140.

1 solution

Boi (보이)
Jun 8, 2018

Since $\displaystyle (n-1)(n-3)a_{n}=\sum_{i=1}^{n-1}a_i,$ we have

$n(n-2)a_{n+1}-(n-1)(n-3)a_n=a_n \\ n(n-2)a_{n+1}=\{(n-1)(n-3)+1\}a_n=(n-2)^2a_n \\ na_{n+1}=(n-2)a_n~(n\neq 2) \\ n(n-1)a_{n+1}=(n-1)(n-2)a_n~(n\neq 2)~{\small \color{#3D99F6} (n-1)(n-2)a_n=b_n} \\ b_{n+1}=b_n=\,\cdots\,= b_3 = 64 \\ a_n=\frac{64}{(n-1)(n-2)}.~(n\ge 3)$

Therefore, we have $\dfrac{1}{a_n}=\dfrac{(n-1)(n-2)}{64}.~(n\ge 3)$

If $\gcd(k-1,~64) \neq 1,$ then $k$ is an odd number, which means $\gcd(k-2,~64)=1.$

Same applies for when $\gcd(k-2,~64) \neq 1.$

Hence, either $k-1$ is a multiple of $64,$ or $k-2$ is a multiple of $64.$

This means $k=1+64m$ or $k=2+64m.$

Note that $k=1$ and $k=2$ also satisfies the given condition, and with $\dfrac{10^{20}}{64}\times 2=3125\times 10^{15},$ we can conclude that

$S(10^{20})=3125\times 10^{15},$

so that $p+q=3125+15=\boxed{3140}.$

#17 of June 2018 Grade 11 - Natural Sciences CSAT (Korean SAT) Mock test

The answer is 3140.

1 solution

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