# 2015 AMC 12B Problems/Problem 18

## Problem

For every composite positive integer $n$, define $r(n)$ to be the sum of the factors in the prime factorization of $n$. For example, $r(50) = 12$ because the prime factorization of $50$ is $2 \times 5^{2}$, and $2 + 5 + 5 = 12$. What is the range of the function $r$, $\{r(n): n \text{ is a composite positive integer}\}$ ? $\textbf{(A)}\; \text{the set of positive integers} \\ \textbf{(B)}\; \text{the set of composite positive integers} \\ \textbf{(C)}\; \text{the set of even positive integers} \\ \textbf{(D)}\; \text{the set of integers greater than 3} \\ \textbf{(E)}\; \text{the set of integers greater than 4}$

## Solution 1

This problem becomes simple once we recognize that the domain of the function is $\{4, 6, 8, 9, 10, 12, 14, 15, \dots\}$. By evaluating $r(4)$ to be $4$, we can see that $\textbf{(E)}$ is incorrect. Evaluating $r(6)$ to be $5$, we see that both $\textbf{(B)}$ and $\textbf{(C)}$ are incorrect. Since our domain consists of composite numbers, which, by definition, are a product of at least two positive primes, the minimum value of $r(n)$ is $4$, so $\textbf{(A)}$ is incorrect. That leaves us with $\boxed{\textbf{(D)}\; \text{the set of integers greater than }3}$.

## Solution 2

Think backwards. The range is the same as the numbers $y$ that can be expressed as the sum of two or more prime positive integers.

The lowest number we can get is $y = 2+2 = 4$. For any number greater than 4, we can get to it by adding some amount of 2's and then possibly a 3 if that number is odd. For example, 23 can be obtained by adding 2 ten times and adding a 3; this corresponds to the argument $n = 2^{10} \times 3$. Thus our answer is $\boxed{\textbf{(D)}\; \text{the set of integers greater than }3}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 