Difference between revisions of "2015 AMC 12B Problems/Problem 18"
Pi over two (talk | contribs) (→Problem) |
Andrew2019 (talk | contribs) (→Solution 4 (Elimination)=) |
||
| (7 intermediate revisions by 4 users not shown) | |||
| Line 2: | Line 2: | ||
For every composite positive integer <math>n</math>, define <math>r(n)</math> to be the sum of the factors in the prime factorization of <math>n</math>. For example, <math>r(50) = 12</math> because the prime factorization of <math>50</math> is <math>2 \times 5^{2}</math>, and <math>2 + 5 + 5 = 12</math>. What is the range of the function <math>r</math>, <math>\{r(n): n \text{ is a composite positive integer}\}</math> ? | For every composite positive integer <math>n</math>, define <math>r(n)</math> to be the sum of the factors in the prime factorization of <math>n</math>. For example, <math>r(50) = 12</math> because the prime factorization of <math>50</math> is <math>2 \times 5^{2}</math>, and <math>2 + 5 + 5 = 12</math>. What is the range of the function <math>r</math>, <math>\{r(n): n \text{ is a composite positive integer}\}</math> ? | ||
| − | <math>\textbf{(A)}\; | + | <math>\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}</math> | ||
==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
Contents
Problem
For every composite positive integer 
, define 
to be the sum of the factors in the prime factorization of . For example, 
because the prime factorization of 
is 
, and 
. What is the range of the function 
, 
?
Solution
Solution 1
This problem becomes simple once we recognize that the domain of the function is 
. By evaluating 
to be 
, we can see that 
is incorrect. Evaluating 
to be 
, we see that both 
and 
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 is , so 
is incorrect. That leaves us with 
.
Solution 2
Think backwards. The range is the same as the numbers 
that can be expressed as the sum of two or more prime positive integers.
The lowest number we can get is 
. 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 
. Thus our answer is .
Solution 3
Notice that by the Chicken McNugget Theorem, given 
s and 
s, we can make any number larger than 
We also exclude and since they correspond to primes. All other numbers can be written as 
which 
will equal. Thus the answer is 
Solution 4 (Elimination)
Note that 
eliminates 
and 
and eliminates 
. We can easily see that cannot be less that by testing and . So, the answer is clearly .
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. 
