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

(Solution 2)
(Solution 2)
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===Solution 2===
 
===Solution 2===
Say a number <math>n</math> is <math>boring</math> if the sequence <math>\{f_k(n)\}</math> is bounded; otherwise <math>n</math> is <math>interesting</math>.  
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Say a number <math>n</math> is <math>\textit{boring}</math> if the sequence <math>\{f_k(n)\}</math> is bounded; otherwise <math>n</math> is <math>\textit{interesting}</math>.  
  
 
Notice that <math>f_1(n)\mid f_1(kn)</math> for all <math>k\in\mathbb{N}</math>; an inductive argument leads to the conclusion that multiples of an interesting number are interesting.  
 
Notice that <math>f_1(n)\mid f_1(kn)</math> for all <math>k\in\mathbb{N}</math>; an inductive argument leads to the conclusion that multiples of an interesting number are interesting.  
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Let <math>S</math> be the subset of <math>[400]</math> consisting of integers whose prime factorization only contains exponents 2 or higher. It suffices to check <math>n\in S</math>: all interesting numbers in <math>[400]</math> are multiples of interesting numbers in <math>S</math>.
 
Let <math>S</math> be the subset of <math>[400]</math> consisting of integers whose prime factorization only contains exponents 2 or higher. It suffices to check <math>n\in S</math>: all interesting numbers in <math>[400]</math> are multiples of interesting numbers in <math>S</math>.
  
Let's first look at all the prime powers <math>>100</math> (these have no other multiples in <math>S</math>). They are: <math>5^3, 7^3, 11^2</math>, <math>13^2</math>, <math>17^2</math>, and <math>19^2</math>. Only '''<math>7^3</math> is interesting''' but it has no other multiples (even in <math>[400]</math>).
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Let's first look at all the prime powers <math>>100</math> (these have no other multiples in <math>S</math>). They are: <math>5^3, 7^3, 11^2</math>, <math>13^2</math>, <math>17^2</math>, and <math>19^2</math>. Only <math>7^3 \textbf{ is interesting}</math> but it has no other multiples (even in <math>[400]</math>).
  
 
We are left with  <math>n\in S</math> such that either <math>5^2\mid n</math> or <math>7^2\mid n</math>.
 
We are left with  <math>n\in S</math> such that either <math>5^2\mid n</math> or <math>7^2\mid n</math>.
  
If <math>7^2\mid n</math> then <math>n</math> is <math>7^2</math>, or <math>2^2\cdot 7^2</math> or <math>2^3\cdot 7^2</math> and all are boring. If <math>5^2\mid n</math> then <math>n=5^2</math>, or <math>2^2\cdot 5^2</math> or <math>n=2^3\cdot 5^2</math> (all three are boring) or <math>n=2^4\cdot 5^2=400</math>, '''which is interesting''', but, of course, has no other multiples.
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If <math>7^2\mid n</math> then <math>n</math> is <math>7^2</math>, or <math>2^2\cdot 7^2</math> or <math>2^3\cdot 7^2</math> and all are boring. If <math>5^2\mid n</math> then <math>n=5^2</math>, or <math>2^2\cdot 5^2</math> or <math>n=2^3\cdot 5^2</math> (all three are boring) or <math>n=2^4\cdot 5^2=400</math>, <math>\textbf{which is interesting}</math>, but, of course, has no other multiples.
  
 
We have <math>12</math> multiples of <math>2^5</math>, <math>4</math> multiples of <math>3^4</math>, <math>2^4\cdot 5^2</math>, and <math>7^3</math> for a total of <math>12+4+1+1= 18</math> ... <math>\framebox{D}</math>.
 
We have <math>12</math> multiples of <math>2^5</math>, <math>4</math> multiples of <math>3^4</math>, <math>2^4\cdot 5^2</math>, and <math>7^3</math> for a total of <math>12+4+1+1= 18</math> ... <math>\framebox{D}</math>.

Revision as of 16:55, 30 October 2021

Problem

Define the function $f_1$ on the positive integers by setting $f_1(1)=1$ and if $n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ is the prime factorization of $n>1$, then \[f_1(n)=(p_1+1)^{e_1-1}(p_2+1)^{e_2-1}\cdots (p_k+1)^{e_k-1}.\] For every $m\ge 2$, let $f_m(n)=f_1(f_{m-1}(n))$. For how many $N$s in the range $1\le N\le 400$ is the sequence $(f_1(N),f_2(N),f_3(N),\dots )$ unbounded?

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

$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19$

Solution

Solution 1

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

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

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

$3^3 \cdot Q^2$ could also work if $Q^2$ satisfies $2~|~f_1(Q^2)$, but $3^3 \cdot 5^2 > 400$.

$f_1(5^3)=6^2 = 2^2 3^2$ does not work.

$f_1(7^3)=8^2=2^6$ works. There are no other multiples of $7^3$ within $400$.

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

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

In conclusion, there are $12+1+4+1=18$ number of $N$'s ... $\framebox{D}$.

Solution 2

Say a number $n$ is $\textit{boring}$ if the sequence $\{f_k(n)\}$ is bounded; otherwise $n$ is $\textit{interesting}$.

Notice that $f_1(n)\mid f_1(kn)$ for all $k\in\mathbb{N}$; an inductive argument leads to the conclusion that multiples of an interesting number are interesting.

We have $f_2(2^a)=f_1(3^{a-1}) = 2^{2(a-2)}$; since $2(a-2)> a$ iff $a\ge 5$, we have $2^5=32$ is the least power of $2$ that is interesting. There are $12$ multiples of $32$ that are $\le 400$.

We have $f_2(3^b)=f_1(2^{2(b-1)}) = 3^{2b-3}$; since $2b-3> b$ iff $b\ge 4$, we have $3^4=81$ is the least power of $3$ that is interesting. There are $4$ multiples of $81$ that are $\le 400$.

We have $f_2(2^a3^b)= 2^{2(a-2)}3^{2b-3}$, so a number of this form is interesting iff $a\ge 5$ or $b\ge 4$; they are already counted above.

Integer $n$ is interesting (resp. boring) iff $f_1(n)$ is. A square-free integer is boring, since $f_1(n) = \prod f_1(p_i)=\prod 1 = 1$.

Let $S$ be the subset of $[400]$ consisting of integers whose prime factorization only contains exponents 2 or higher. It suffices to check $n\in S$: all interesting numbers in $[400]$ are multiples of interesting numbers in $S$.

Let's first look at all the prime powers $>100$ (these have no other multiples in $S$). They are: $5^3, 7^3, 11^2$, $13^2$, $17^2$, and $19^2$. Only $7^3 \textbf{ is interesting}$ but it has no other multiples (even in $[400]$).

We are left with $n\in S$ such that either $5^2\mid n$ or $7^2\mid n$.

If $7^2\mid n$ then $n$ is $7^2$, or $2^2\cdot 7^2$ or $2^3\cdot 7^2$ and all are boring. If $5^2\mid n$ then $n=5^2$, or $2^2\cdot 5^2$ or $n=2^3\cdot 5^2$ (all three are boring) or $n=2^4\cdot 5^2=400$, $\textbf{which is interesting}$, but, of course, has no other multiples.

We have $12$ multiples of $2^5$, $4$ multiples of $3^4$, $2^4\cdot 5^2$, and $7^3$ for a total of $12+4+1+1= 18$ ... $\framebox{D}$.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc12b/276

~dolphin7


See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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