# Difference between revisions of "2012 AMC 12B Problems/Problem 24"

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− | First of all, notice that for any odd prime <math>p</math>, the largest prime that divides <math>p+1</math> is no larger than <math>\frac{p+1}{2}</math>, therefore eventually the factorization of <math>f_k(N)</math> does not contain any prime larger than <math>3</math>. Also, note that <math>f_2(2^m) = f_1(3^{m-1})=2^{2m-4}</math>, when <math>m=4</math> it stays the same but when <math>m\geq 5</math> it grows indefinitely. Therefore any number <math>N</math> that is divisible by <math>2^5</math> or any number <math>N</math> such that <math>f_k(N)</math> is divisible by <math>2^5</math> makes the sequence <math>(f_1(N),f_2(N),f_3(N),\dots )</math> unbounded. There are <math>12</math> multiples of <math>2^5</math> within <math>400</math>. | + | First of all, notice that for any odd prime <math>p</math>, the largest prime that divides <math>p+1</math> is no larger than <math>\frac{p+1}{2}</math>, therefore eventually the factorization of <math>f_k(N)</math> does not contain any prime larger than <math>3</math>. Also, note that <math>f_2(2^m) = f_1(3^{m-1})=2^{2m-4}</math>, when <math>m=4</math> it stays the same but when <math>m\geq 5</math> it grows indefinitely. Therefore any number <math>N</math> that is divisible by <math>2^5</math> or any number <math>N</math> such that <math>f_k(N)</math> is divisible by <math>2^5</math> makes the sequence <math>(f_1(N),f_2(N),f_3(N),\dots )</math> unbounded. There are <math>12</math> multiples of <math>2^5</math> within <math>400</math>. <math>2^4 5^2=400</math> also works: <math>f_2(2^4 5^2) = f_1(3^4 \cdot 2) = 2^6</math>. |

− | Any first power of prime in a prime factorization will not contribute the unboundedness because <math>f_1(p^1)=(p+1)^0=1</math>. At least a square of prime is to contribute. So we test primes that are less than <math>\sqrt{400}=20</math>: | + | Now let's look at the other cases. Any first power of prime in a prime factorization will not contribute the unboundedness because <math>f_1(p^1)=(p+1)^0=1</math>. At least a square of prime is to contribute. So we test primes that are less than <math>\sqrt{400}=20</math>: |

<math>f_1(3^4)=4^3=2^6</math> works, therefore any number <math>\leq 400</math> that are divisible by <math>3^4</math> works: there are <math>4</math> of them. | <math>f_1(3^4)=4^3=2^6</math> works, therefore any number <math>\leq 400</math> that are divisible by <math>3^4</math> works: there are <math>4</math> of them. | ||

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<math>7^2 \cdot Q^2</math> could also work if <math>4~|~f_1(Q^2)</math>, but <math>7^2 \cdot 3^2 > 400</math> already. | <math>7^2 \cdot Q^2</math> could also work if <math>4~|~f_1(Q^2)</math>, but <math>7^2 \cdot 3^2 > 400</math> already. | ||

− | For number that are only divisible by <math>p=11, 13, 19</math>, they don't work because none of these primes are such that <math>p+1</math> could be a multiple of <math>2^5</math> nor a multiple of <math>3^4</math>. | + | For number that are only divisible by <math>p=11, 13, 17, 19</math>, they don't work because none of these primes are such that <math>p+1</math> could be a multiple of <math>2^5</math> nor a multiple of <math>3^4</math>. |

− | + | In conclusion, there are <math>12+1+4+1=18</math> number of <math>N</math>'s ... <math>\framebox{D}</math>. | |

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− | In conclusion, there are <math>12+4 |

## Revision as of 04:55, 6 December 2012

## Problem 24

Define the function on the positive integers by setting and if is the prime factorization of , then For every , let . For how many in the range is the sequence unbounded?

**Note:** A sequence of positive numbers is unbounded if for every integer , there is a member of the sequence greater than .

## Solution

First of all, notice that for any odd prime , the largest prime that divides is no larger than , therefore eventually the factorization of does not contain any prime larger than . Also, note that , when it stays the same but when it grows indefinitely. Therefore any number that is divisible by or any number such that is divisible by makes the sequence unbounded. There are multiples of within . also works: .

Now let's look at the other cases. Any first power of prime in a prime factorization will not contribute the unboundedness because . At least a square of prime is to contribute. So we test primes that are less than :

works, therefore any number that are divisible by works: there are of them.

could also work if satisfies , but .

does not work.

works. There are no other multiples of within .

could also work if , but already.

For number that are only divisible by , they don't work because none of these primes are such that could be a multiple of nor a multiple of .

In conclusion, there are number of 's ... .