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

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Let the prime factorization of <math>x_i</math> be written as <math>p_{a_1} \cdot p_{a_2} \cdot p_{a_3} \cdots</math>, where <math>p_i</math> is the <math>i</math>th prime number. Then, for every <math>p_{a_k}</math> in the prime factorization of <math>x_i</math>, place a <math>1</math> in the <math>a_k</math>th digit of <math>y_i</math>. This will result in the conversion <math>x_1 = 2, x_{2} = 3, x_{3} = 2 * 3 = 6, \cdots</math>.
 
Let the prime factorization of <math>x_i</math> be written as <math>p_{a_1} \cdot p_{a_2} \cdot p_{a_3} \cdots</math>, where <math>p_i</math> is the <math>i</math>th prime number. Then, for every <math>p_{a_k}</math> in the prime factorization of <math>x_i</math>, place a <math>1</math> in the <math>a_k</math>th digit of <math>y_i</math>. This will result in the conversion <math>x_1 = 2, x_{2} = 3, x_{3} = 2 * 3 = 6, \cdots</math>.
  
−
Multiplication for the sequence <math>x_i</math> will translate to addition for the sequence <math>y_i</math>. Thus, we see that <math>x_{n+1}X(x_n) = x_np(x_n)</math> translates into <math>y_{n+1} = y_n+1</math>. Since <math>x_0=1, and y_0=0</math>, <math>x_i</math> corresponds to <math>y_i</math>, which is <math>i</math> in binary. Since <math>x_{10010101_2} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090</math>, <math>t = 10010101_2</math> = <math>\boxed{149}</math>.
+
Multiplication for the sequence <math>x_i</math> will translate to addition for the sequence <math>y_i</math>. Thus, we see that <math>x_{n+1}X(x_n) = x_np(x_n)</math> translates into <math>y_{n+1} = y_n+1</math>. Since <math>x_0=1</math>, and<math> y_0=0</math>, <math>x_i</math> corresponds to <math>y_i</math>, which is <math>i</math> in binary. Since <math>x_{10010101_2} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090</math>, <math>t = 10010101_2</math> = <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 21:05, 10 January 2016

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 $x_{10010101_2} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090$, $t = 10010101_2$ = $\boxed{149}$.

See also

2014 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Problem
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All AIME Problems and Solutions

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