Difference between revisions of "2021 Fall AMC 12B Problems/Problem 12"
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The prime factorization of <math>768</math> is <math>2^8\cdot3,</math> and the prime factorization of <math>384</math> is <math>2^7\cdot3.</math> Note that | The prime factorization of <math>768</math> is <math>2^8\cdot3,</math> and the prime factorization of <math>384</math> is <math>2^7\cdot3.</math> Note that | ||
<cmath>\begin{alignat*}{8} | <cmath>\begin{alignat*}{8} | ||
− | f(768) &= \left(\frac{2^9-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div768 &&= \frac{ | + | f(768) &= \left(\frac{2^9-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div768 &&= \frac{2044}{768}, \\ |
− | f(384) &= \left(\frac{2^8-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div384 &&= \frac{ | + | f(384) &= \left(\frac{2^8-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div384 &&= \frac{1020}{384}. |
\end{alignat*}</cmath> | \end{alignat*}</cmath> | ||
− | Therefore, the answer is < | + | Therefore, the answer is <math>f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.</math> |
+ | |||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
==Solution 3== | ==Solution 3== | ||
− | 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> | + | 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> |
− | |||
− | 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> | ||
− | Finally, the answer is < | + | Finally, the answer is <math>f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.</math> |
~mahaler | ~mahaler |
Revision as of 02:03, 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
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
Solution 3
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
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.