Difference between revisions of "2021 Fall AMC 12B Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
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+ | The prime factorization of <math>384</math> is <math>2^7\cdot3,</math> so each of its positive divisors is in the form <math>2^m</math> or <math>2^m\cdot3</math> for some nonnegative integer <math>m\leq7.</math> We will use this fact to calculate the sum of all its positive divisors. Note that <cmath>2^m + 2^m\cdot3 = 2^m\cdot(1+3) = 2^m\cdot4 = 2^m\cdot2^2 = 2^{m+2}</cmath> is the sum of the two forms of positive divisors for each <math>m</math> from <math>0</math> through <math>7.</math> Therefore, the sum of all positive divisors of <math>384</math> is <cmath>2^2 + 2^3 + 2^4 + \cdots + 2^9 = \frac{2^{10}-2^2}{2-1} = 1020,</cmath> from which <math>f(384) = \frac{1020}{384}.</math> Similarly, since the prime factorization of <math>768</math> is <math>2^8 \cdot 3,</math> we have <math>f(768) = \frac{2044}{768}.</math> | ||
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+ | Finally, the answer is <math>f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.</math> | ||
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+ | ~mahaler | ||
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+ | ==Solution 3== | ||
Let <math>\sigma(n)</math> denotes the sum of the positive integer divisors of <math>n,</math> so <math>f(n)=\frac{\sigma(n)}{n}.</math> | Let <math>\sigma(n)</math> denotes the sum of the positive integer divisors of <math>n,</math> so <math>f(n)=\frac{\sigma(n)}{n}.</math> | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2021 Fall|ab=B|num-a=13|num-b=11}} | {{AMC12 box|year=2021 Fall|ab=B|num-a=13|num-b=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:05, 27 January 2022
Problem
For a positive integer, let be the quotient obtained when the sum of all positive divisors of is divided by . For example, What is
Solution 1
The prime factorization of is and the prime factorization of is Note that Therefore, the answer is
~lopkiloinm ~MRENTHUSIASM
Solution 2
The prime factorization of is so each of its positive divisors is in the form or for some nonnegative integer We will use this fact to calculate the sum of all its positive divisors. Note that is the sum of the two forms of positive divisors for each from through Therefore, the sum of all positive divisors of is from which Similarly, since the prime factorization of is we have
Finally, the answer is
~mahaler
Solution 3
Let denotes the sum of the positive integer divisors of so
Suppose that is the prime factorization of Since is multiplicative, we have The prime factorization of is and the prime factorization of is Note that Therefore, the answer is
~MRENTHUSIASM
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.