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

(Solution)
Line 3: Line 3:
  
 
==Solution==
 
==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 * 5 * 11 * 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, ..., a_{10010101} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090.</math> Thus, <math>t = 10010101</math> (base 2) = <math>\boxed{149}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=II|num-b=14|after=Last Problem}}
 
{{AIME box|year=2014|n=II|num-b=14|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:46, 20 April 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) $a_1 = 2, a_{10} = 3, a_{11} = 2 * 3 = 6, ..., a_{10010101} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090.$ Thus, $t = 10010101$ (base 2) = $\boxed{149}$.

See also

2014 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png