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

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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 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 $\textit{interesting}$ .

It is clear that $n$ is interesting iff $f_1(n)$ is. For a prime, $p$ , $f_1(p)=1$ . Thus, for a prime, $p_k$ , in the prime factorization of $n$ ; if $e_k=1$ then $f_1(n)=f_1(n/p_k)$ , so $n$ is interesting iff $n/p_k$ is. Continuing in this manner, we can divide $n$ by all such primes $p_k$ for which $e_k=1$ ; and $n$ is interesting iff each of these resulting numbers are. Finally we will end up with a number whose prime factorization contains only exponents $\ge 2$ . Let $S$ be the set of such numbers. It suffices to find all interesting numbers in $S$ ; all other interesting numbers will be multiples of these. Note that $f_1(n)\mid f_1(kn)$ for all $k\in\mathbb{N}$ ; so by induction a multiple of an interesting number is also interesting.

The set $S$ contains either powers of primes $\le 19$ , or a product of two such powers.

We have $f_2(2^a)=f_1(3^{a-1}) = 2^{2(a-2)}$ ; since $2(a-2)> a$ iff $a\ge 5$ , so $2^5$ and its multiples are interesting; there are $12$ such.

We have $f_2(3^b)=f_1(2^{2(b-1)}) = 3^{2b-3}$ ; since $2b-3> b$ iff $b\ge 4$ , so $3^4$ and its multiples are interesting; there are $4$ such.

Among the remaining prime powers only $7^3 \textbf{ is interesting}$ but it has no other multiples in $[400]$ .

We are now left with numbers $n\in S$ that are products of two prime powers, i.e $n=p^aq^b$ , $a,b \ge 2$ .

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.

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

If $7^2\mid n$ then $n$ is $2^2\cdot 7^2$ or $2^3\cdot 7^2$ and both are boring.

No larger prime can divide $n$ because even $11^2\cdot 2^2 > 400$ .

We have found all the interesting numbers -- 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

@@ Line 49: / Line 49: @@
 If <math>7^2\mid n</math> then <math>n</math> is <math>2^2\cdot 7^2</math> or <math>2^3\cdot 7^2</math> and both are boring.
-No larger prime can divide <math>n</math> because even <math>11^2\cdot 2^2 > 400</math>. We have found all the interesting numbers.
+No larger prime can divide <math>n</math> because even <math>11^2\cdot 2^2 > 400</math>.
-We have 12 multiples of <math>2^5</math>, 4 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 found all the interesting numbers -- 12 multiples of <math>2^5</math>, 4 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>.
 ==Video Solution by Richard Rusczyk==

2012 AMC 12B (Problems • Answer Key • Resources)
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