Difference between revisions of "2015 AMC 12B Problems/Problem 18"
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==Solution== | ==Solution== | ||
− | + | 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>. | |
==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}} |
Revision as of 11:28, 5 March 2015
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
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 .
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 |
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