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

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<math>\textbf{-RootThreeOverTwo}</math>
 
<math>\textbf{-RootThreeOverTwo}</math>
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==Solution 3 (induction)==
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Let <math>P_k</math> denote the <math>k</math>th prime.
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Lemma: <math>x_{n+2^{k-1}} = P_k \cdot x_{n}</math> for all <math>0 \leq n \leq 2^{k-1} - 1.</math>
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[b]Proof:[/b]
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We can prove this using induction. Assume that <math>x_{2^{k-1}-1} = \prod_{j=1}^{k-1} P_j. </math> Then, using the given recursion <math>x_{k+1} = \frac{x_np(x_n)}{X(x_n}</math>, we would “start fresh” for <math>x_{2^{k-1}} = P_k.</math> It is then easy to see that then <math>\frac{x_n}{P_k}</math> just cycles through the previous <math>x_{2^{k-1}}</math> terms of <math>\{ x_n \}, </math> since the recursion process is the same and <math>P_k</math> being a factor of <math>x_n</math> is not affected until <math>n = 2 \cdot {2^{k-1}} = 2^k, </math> when given our assumption <math>x_{2^k - 1} = \prod_{j=1}^{k} P_j</math> and <math>n = 2^k</math> is now the least <math>n</math> such that <cmath>P_{k+1} = p(x_{2^{n-1}}), </cmath> in which <math>P_a = p(x_n)</math> where <math>a > k</math> is the only way that the aforementioned cycle would be affected. Specifically, by cancellation according to our recursion, <math>x_{2^k} = P_{k+1}, </math> and the values of <math>x_n</math> just starts cycling through the previous <math>x_{2^k}</math> terms again until <math>x_{2^{k+1}}</math> when a new prime shows up in the prime factorization of <math>x_n, </math> when it starts cycling again, and so on. Using our base cases of <math>x_0</math> and <math>x_1, </math> our induction is complete.
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Now, it is easy to see that <math>2090 = 2 \cdot 5 \cdot 11 \cdot 19 = P_1 \cdot P_3 \cdot P_5 \cdot P_8,</math> and by Lemma #1, the least positive integer <math>n</math> such that <math>19 | x_n</math> is <math>2^7. </math> By repeatedly applying our obtained recursion from Lemma #1, it is easy to see that our answer is just <math>2^7 + 2^4 + 2^2 + 2^0, </math> or <math>10010101 = \boxed{149}_2.</math>
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-fidgetboss_4000
  
 
== 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 11:21, 12 May 2020

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

Solution 2 (Painful Bash)

We go through the terms and look for a pattern. We find that

$x_0 = 1$ $x_8 = 7$

$x_1 = 2$ $x_9 = 14$

$x_2 = 3$ $x_{10} = 21$

$x_3 = 6$ $x_{11} = 42$

$x_4 = 5$ $x_{12} = 35$

$x_5 = 10$ $x_{13} = 70$

$x_6 = 15$ $x_{14} = 105$

$x_7 = 30$ $x_{15} = 210$

Commit to the bash. Eventually, you will receive that $x_{149} = 2090$, so $\boxed{149}$ is the answer. Trust me, this is worth the 10 index points.

$\textbf{-RootThreeOverTwo}$

Solution 3 (induction)

Let $P_k$ denote the $k$th prime. Lemma: $x_{n+2^{k-1}} = P_k \cdot x_{n}$ for all $0 \leq n \leq 2^{k-1} - 1.$ [b]Proof:[/b] We can prove this using induction. Assume that $x_{2^{k-1}-1} = \prod_{j=1}^{k-1} P_j.$ Then, using the given recursion $x_{k+1} = \frac{x_np(x_n)}{X(x_n}$, we would “start fresh” for $x_{2^{k-1}} = P_k.$ It is then easy to see that then $\frac{x_n}{P_k}$ just cycles through the previous $x_{2^{k-1}}$ terms of $\{ x_n \},$ since the recursion process is the same and $P_k$ being a factor of $x_n$ is not affected until $n = 2 \cdot {2^{k-1}} = 2^k,$ when given our assumption $x_{2^k - 1} = \prod_{j=1}^{k} P_j$ and $n = 2^k$ is now the least $n$ such that \[P_{k+1} = p(x_{2^{n-1}}),\] in which $P_a = p(x_n)$ where $a > k$ is the only way that the aforementioned cycle would be affected. Specifically, by cancellation according to our recursion, $x_{2^k} = P_{k+1},$ and the values of $x_n$ just starts cycling through the previous $x_{2^k}$ terms again until $x_{2^{k+1}}$ when a new prime shows up in the prime factorization of $x_n,$ when it starts cycling again, and so on. Using our base cases of $x_0$ and $x_1,$ our induction is complete. Now, it is easy to see that $2090 = 2 \cdot 5 \cdot 11 \cdot 19 = P_1 \cdot P_3 \cdot P_5 \cdot P_8,$ and by Lemma #1, the least positive integer $n$ such that $19 | x_n$ is $2^7.$ By repeatedly applying our obtained recursion from Lemma #1, it is easy to see that our answer is just $2^7 + 2^4 + 2^2 + 2^0,$ or $10010101 = \boxed{149}_2.$

-fidgetboss_4000

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

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