Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
(→‎Solution 4 (Elimination)=)
 
(6 intermediate revisions by 4 users not shown)
Line 10: Line 10:
 
==Solution==
 
==Solution==
  
 +
==Solution 1==
 +
 +
This problem becomes simple once we recognize that the domain of the function is <math>\{4, 6, 8, 9, 10, 12, 14, 15, \dots\}</math>. By evaluating <math>r(4)</math> to be <math>4</math>, we can see that <math>\textbf{(E)}</math> is incorrect. Evaluating <math>r(6)</math> to be <math>5</math>, we see that both <math>\textbf{(B)}</math> and <math>\textbf{(C)}</math> 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 <math>r(n)</math> is <math>4</math>, so <math>\textbf{(A)}</math> is incorrect. That leaves us with <math>\boxed{\textbf{(D)}\; \text{the set of integers greater than }3}</math>.
 +
 +
==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 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>.
 +
 +
==Solution 3==
 +
Notice that by the Chicken McNugget Theorem, given <math>2</math>s and <math>3</math>s, we can make any number larger than <math>2\cdot3 - 2 - 3 = 1.</math> We also exclude <math>2</math> and <math>3</math> since they correspond to primes. All other numbers can be written as <math>2a + 3b,</math> which <math>r(2^a3^b)</math> will equal. Thus the answer is <math>D.</math>
 +
 +
==Solution 4 (Elimination)==
 +
Note that <math>r(10)</math> eliminates <math>B</math> and <math>C</math> and <math>r(4)</math> eliminates <math>E</math>. We can easily see that <math>r(n)</math> cannot be less that <math>3</math> by testing <math>2</math> and <math>3</math>. So, the answer is clearly <math>\boxed{\textbf{(D)}\; \text{the set of integers greater than }3}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=19|num-b=17}}
 
{{AMC12 box|year=2015|ab=B|num-a=19|num-b=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:14, 2 June 2024

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

Solution 3

Notice that by the Chicken McNugget Theorem, given $2$s and $3$s, we can make any number larger than $2\cdot3 - 2 - 3 = 1.$ We also exclude $2$ and $3$ since they correspond to primes. All other numbers can be written as $2a + 3b,$ which $r(2^a3^b)$ will equal. Thus the answer is $D.$

Solution 4 (Elimination)

Note that $r(10)$ eliminates $B$ and $C$ and $r(4)$ eliminates $E$. We can easily see that $r(n)$ cannot be less that $3$ by testing $2$ and $3$. So, the answer is clearly $\boxed{\textbf{(D)}\; \text{the set of integers greater than }3}$.

See Also

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_18&oldid=220001"