# 2014 AIME II Problems/Problem 15

## Problem

For any integer $k\geq 1$, let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$, and $X(k)=1$ if $p(k)=2.$ Let $\{x_n\}$ be the sequence defined by $x_0=1$, and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0.$ Find the smallest positive integer $t$ such that $x_t=2090.$

## Solution

Note that for any $x_i$, for any prime $p$, $p^2 \nmid x_i$. This provides motivation to translate $x_i$ into a binary sequence $y_i$.

Let the prime factorization of $x_i$ be written as $p_{a_1} \cdot p_{a_2} \cdot p_{a_3} \cdots$, where $p_i$ is the $i$th prime number. Then, for every $p_{a_k}$ in the prime factorization of $x_i$, place a $1$ in the $a_k$th digit of $y_i$. This will result in the conversion $x_1 = 2, x_{2} = 3, x_{3} = 2 * 3 = 6, \cdots$.

Multiplication for the sequence $x_i$ will translate to addition for the sequence $y_i$. Thus, we see that $x_{n+1}X(x_n) = x_np(x_n)$ translates into $y_{n+1} = y_n+1$. Since $x_0=1$, and $y_0=0$, $x_i$ corresponds to $y_i$, which is $i$ in binary. Since $x_{10010101_2} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090$, $t = 10010101_2$ = $\boxed{149}$.

## Solution 2 (Painful Bash)

We go through the terms and look for a pattern. We find that

$x_0 = 1$ $x_8 = 7$

$x_1 = 2$ $x_9 = 14$

$x_2 = 3$ $x_{10} = 21$

$x_3 = 6$ $x_{11} = 42$

$x_4 = 5$ $x_{12} = 35$

$x_5 = 10$ $x_{13} = 70$

$x_6 = 15$ $x_{14} = 105$

$x_7 = 30$ $x_{15} = 210$

Commit to the bash. Eventually, you will recieve that $x_{149} = 2090$, so $\boxed{149}$ is the answer. Trust me, this is worth the 10 index points.

$\textbf{-RootThreeOverTwo}$