Difference between revisions of "2021 Fall AMC 12B Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
− | The prime factorization of <math>384</math> is <math>2^7\cdot3,</math> so each of its positive divisors is of the form <math>2^m</math> or <math>2^m\cdot3</math> for some integer <math>m</math> such that <math>0\leq 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 all such <math>m.</math> By geometric series, the sum of all positive divisors of <math>384</math> is <cmath>\sum_{k=2}{9}2^k = \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> | + | The prime factorization of <math>384</math> is <math>2^7\cdot3,</math> so each of its positive divisors is of the form <math>2^m</math> or <math>2^m\cdot3</math> for some integer <math>m</math> such that <math>0\leq 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 all such <math>m.</math> By geometric series, the sum of all positive divisors of <math>384</math> is <cmath>\sum_{k=2}^{9}2^k = \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> |
Therefore, the answer is <math>f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.</math> | Therefore, the answer is <math>f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.</math> |
Revision as of 03:07, 18 February 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
is the sum of all fractions of the form
where
is a positive divisor of
By geometric series, it follows that
Therefore, the answer is
~lopkiloinm ~MRENTHUSIASM
Solution 2
The prime factorization of is
so each of its positive divisors is of the form
or
for some integer
such that
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 all such
By geometric series, the sum of all positive divisors of
is
from which
Similarly, since the prime factorization of
is
we have
Therefore, the answer is
~mahaler
Solution 3
Let denotes the sum of all positive divisors of
so
Suppose that is the prime factorization of
Since
is multiplicative, we have
by geometric series.
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.