2014 AIME II Problems/Problem 15

Revision as of 15:10, 3 July 2015 by Ryanyz10 (talk | contribs) (Solution)

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 (in base 2 for the indices, base 10 for the values) $a_1 = 2, a_{10} = 3, a_{11} = 2 * 3 = 6, \cdots, a_{10010101} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090.$ Thus, $t = 10010101$ (base 2) = $\boxed{149}$.

Could someone provide a little more detail? This solution doesn't explain how it got to the answer very well.

See also

2014 AIME II (ProblemsAnswer KeyResources)
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Problem 14
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