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 $x_{10010101} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090,$ t = 10010101 $(base 2) =$ \boxed{149}$.

2014 AIME II (Problems • Answer Key • Resources)
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2014 AIME II Problems/Problem 15

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