Difference between revisions of "2021 Fall AMC 12B Problems/Problem 12"
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so the difference is <math>\boxed{\textbf{(B)}\ \frac{1}{192}}</math> | so the difference is <math>\boxed{\textbf{(B)}\ \frac{1}{192}}</math> | ||
~lopkiloinm | ~lopkiloinm | ||
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+ | ==Solution 2== | ||
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+ | We see that the prime factorization of <math>384</math> is <math>2^7 \cdot 3</math>. Each of its divisors is in the form of <math>2^x</math> or <math>2^x \cdot 3</math> for a nonnegative integer <math>x \le 7</math>. We can use this fact to our advantage when calculating the sum of all of them. Notice that <math>2^x + 2^x \cdot 3 = 2^x(1+3) = 2^x \cdot 4 = 2^x \cdot 2^2 = 2^{x+2}</math> is the sum of the two forms of divisors for each <math>x</math> from <math>0-7</math>, inclusive. So, the sum of all of the divisors of <math>384</math> is just <math>2^2 + 2^3 + 2^4 + \ldots + 2^9 = (2^0 + 2^1 + 2^2 + 2^3 + 2^4 + \ldots + 2^9) - (2^0 + 2^1) = (2^{10} - 1) - (2^0 + 2^1) = 1020</math>. Therefore, <math>f(384) = \frac{1020}{384}</math>. Similarly, since <math>768 = 2^8 \cdot 3</math>, <math>f(768) = \frac{2044}{768}</math>. Therefore, the answer is <math>f(768) - f(384) = \frac{2044}{768} - \frac{1020}{384} = \frac{2044}{768} - \frac{2040}{768} = \frac{4}{768} = \boxed{\textbf{(B)}\ \frac{1}{192}}</math>. | ||
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+ | ~mahaler | ||
==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 22:26, 12 January 2022
Contents
Problem
For a positive integer, let be the quotient obtained when the sum of all positive divisors of n is divided by n. For example, What is
Solution 1
The prime factorization of is and the prime factorization of is so so the difference is ~lopkiloinm
Solution 2
We see that the prime factorization of is . Each of its divisors is in the form of or for a nonnegative integer . We can use this fact to our advantage when calculating the sum of all of them. Notice that is the sum of the two forms of divisors for each from , inclusive. So, the sum of all of the divisors of is just . Therefore, . Similarly, since , . Therefore, the answer is .
~mahaler
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 |
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