Difference between revisions of "2015 AMC 12B Problems/Problem 18"

Revision as of 14:35, 5 March 2015

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

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 $\boxed{\textbf{(D)}\; \text{the set of integers greater than }3}$ .

@@ Line 15: / Line 15: @@
 ==Solution 2==
 Think backwards. The range is the same as the numbers <math>y</math> that can be expressed as the sum of two or more prime positive integers.
-The lowest number we can get is <math>y = 2+2 = 4</math>. For any number greater than <math>4</math>, we can get to it by adding some amount of <math>2</math>'s and then possibly a <math>3</math> if that number is odd. For example, <math>23</math> can be obtained by adding <math>2</math> 10 times and adding a <math>3</math>; this corresponds to the argument <math>n = 2^{10 } \times 3</math>.  Thus our answer is <math>\boxed{\textbf{(D)}\; \text{the set of integers greater than }3}</math>.
+The lowest number we can get is <math>y = 2+2 = 4</math>. 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 <math>n = 2^{10} \times 3</math>.  Thus our answer is <math>\boxed{\textbf{(D)}\; \text{the set of integers greater than }3}</math>.
 ==See Also==
 {{AMC12 box|year=2015|ab=B|num-a=19|num-b=17}}
 {{MAA Notice}}

2015 AMC 12B (Problems • Answer Key • Resources)
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