Difference between revisions of "2014 AIME II Problems/Problem 15"

Revision as of 14:09, 3 July 2015

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.**

@@ Line 3: / Line 3: @@
 ==Solution==
-Note that (in base 2 for the indices, base 10 for the values) <math>a_1 = 2, a_{10} = 3, a_{11} = 2 * 3 = 6, ..., a_{10010101} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090.</math> Thus, <math>t = 10010101</math> (base 2) = <math>\boxed{149}</math>.
+Note that (in base 2 for the indices, base 10 for the values) <math>a_1 = 2, a_{10} = 3, a_{11} = 2 * 3 = 6, \cdots, a_{10010101} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090.</math> Thus, <math>t = 10010101</math> (base 2) = <math>\boxed{149}</math>.
+**Could someone provide a little more detail. This solution doesn't explain how it got to the answer very well.**
 == See also ==
 {{AIME box|year=2014|n=II|num-b=14|after=Last Problem}}
 {{MAA Notice}}

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Difference between revisions of "2014 AIME II Problems/Problem 15"

Revision as of 14:09, 3 July 2015

Problem

Solution

See also