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(1+\frac{1}{2}+\ | + | f(768)&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^8}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2^8\cdot3}\right)&&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^8}\right)\left(1+\frac{1}{3}\right)&&=\frac{511}{256}\cdot\frac{4}{3}&&=\frac{511}{192}, \\ |
− | f(384)&=\left(1+\frac{1}{2}+\ | + | f(384)&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^7}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2^7\cdot3}\right)&&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^7}\right)\left(1+\frac{1}{3}\right)&&=\frac{255}{128}\cdot\frac{4}{3}&&=\frac{85}{32}. |
\end{alignat*}</cmath> | \end{alignat*}</cmath> | ||
− | Therefore, the answer is < | + | Therefore, the answer is <cmath>f(768)-f(384)=\frac{511}{192}-\frac{85}{32}=\frac{511}{192}-\frac{510}{192}=\boxed{\textbf{(B)}\ \frac{1}{192}}.</cmath> |
− | |||
~lopkiloinm ~MRENTHUSIASM | ~lopkiloinm ~MRENTHUSIASM | ||
==Solution 2== | ==Solution 2== | ||
− | We see that the prime factorization of <math>384</math> is <math>2^7 \cdot 3</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> |
+ | |||
+ | Suppose that <math>n=\prod_{i=1}^{k}p_i^{e_i}</math> is the prime factorization of <math>n.</math> Since <math>\sigma(n)</math> is multiplicative, we have <cmath>\sigma(n)=\sigma\left(\prod_{i=1}^{k}p_i^{e_i}\right)=\prod_{i=1}^{k}\sigma\left(p_i^{e_i}\right)=\prod_{i=1}^{k}\left(\sum_{j=0}^{e_i}p_i^j\right)=\prod_{i=1}^{k}\frac{p_i^{e_i+1}-1}{p_i-1}.</cmath> | ||
+ | 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} | ||
+ | f(768) &= \left(\frac{2^9-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div768 &&= \frac{511}{192}, \\ | ||
+ | f(384) &= \left(\frac{2^8-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div384 &&= \frac{85}{32}. | ||
+ | \end{alignat*}</cmath> | ||
+ | Therefore, the answer is <cmath>f(768)-f(384)=\frac{511}{192}-\frac{85}{32}=\frac{511}{192}-\frac{510}{192}=\boxed{\textbf{(B)}\ \frac{1}{192}}.</cmath> | ||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | 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 <cmath>2^x + 2^x \cdot 3 = 2^x(1+3) = 2^x \cdot 4 = 2^x \cdot 2^2 = 2^{x+2}</cmath> is the sum of the two forms of divisors for each <math>x</math> from <math>0</math> to <math>7,</math> inclusive. So, the sum of all of the divisors of <math>384</math> is just <cmath>2^2 + 2^3 + 2^4 + \cdots + 2^9 = (2^0 + 2^1 + 2^2 + 2^3 + 2^4 + \cdots + 2^9) - (2^0 + 2^1) = (2^{10} - 1) - (2^0 + 2^1) = 1020.</cmath> Therefore, we have <math>f(384) = \frac{1020}{384}.</math> Similarly, since <math>768 = 2^8 \cdot 3,</math> we have <math>f(768) = \frac{2044}{768}.</math> | ||
+ | |||
+ | Finally, the answer is <cmath>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}}.</cmath> | ||
~mahaler | ~mahaler |
Revision as of 19:09, 26 January 2022
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
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
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
to
inclusive. So, the sum of all of the divisors of
is just
Therefore, we have
Similarly, since
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.