Difference between revisions of "2021 Fall AMC 12B Problems/Problem 12"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | ==Solution 4== | ||
+ | Note that | ||
+ | <cmath>\begin{align*} | ||
+ | 384 &= 2^7\cdot3, \\ | ||
+ | 768 &= 2^8\cdot3. | ||
+ | \end{align*}</cmath> | ||
+ | Both numbers take the form <math>2^n\cdot3</math>. Define a function <math>S(n)</math> as being the sum of the factors for any number <math>2^n\cdot3</math>, where <math>n</math> is a nonnegative integer. | ||
− | = | + | If you list out the factors of a number <math>2^n\cdot3</math>, all the factors are even except for <math>1</math> and <math>3</math>. So to produce a list of factors for the number <math>2^{n+1}\cdot3</math>, you can multiply all the factors on the first list by <math>2</math> and then add on <math>1</math> and <math>3</math>. Thus, <math>S(n) = 2S(n-1) + 4</math>. |
− | <math>384 = | + | We need to find <math>\frac{S(8)}{768} - \frac{S(7)}{384}</math>. This can be written as <math>\frac{2S(7)+4-2S(7)}{768}</math>. Our recursive formula states that <math>S(n) = 2S(n-1) + 4</math> so <math>S(n) - 2S(n-1) = 4</math>. Thus, the fraction simplifies to <math>\frac{4}{768} = \boxed{\textbf{(B)}\ \frac{1}{192}}</math>. Note that we never needed to figure out what <math>S(8)</math> or <math>S(7)</math> were, sparing us a lot of calculation. |
− | + | ~Curious_crow | |
+ | ==Solution 5== | ||
+ | We know that | ||
+ | <cmath>\begin{align*} | ||
+ | f(768) &= \frac{1 + 2 + 3 + 4 + \cdots + 768}{768}, \\ | ||
+ | f(384) &= \frac{1 + 2 + 3 + 4 + \cdots + 384}{384}. | ||
+ | \end{align*}</cmath> | ||
+ | We want to find the difference of these so | ||
+ | <cmath>\frac{1 + 2 + 3 + 4 + \cdots + 768}{768} - \frac{ 2(1 + 2 + 3 + 4 + \cdots + 384)}{2\cdot384} = \frac{1 + 2 + 3 + 4 + \cdots + 768}{768} - \frac{ 2 + 4 + 6 + 8 + \cdots + 768}{768}.</cmath> | ||
+ | We are only left with the even divisors of <math>768</math> on the fraction on the right so the answer would be the sum of all odd divisors of <math>768</math> (<math>1</math> and <math>3</math>) over <math>768</math>: | ||
+ | <cmath>\frac{1 + 3}{768}=\boxed{\textbf{(B)}\ \frac{1}{192}}.</cmath> | ||
+ | ~pengf | ||
− | + | ==Solution 6== | |
+ | As in Solution 1, the prime factorization of <math>768</math> is <math>2^8 \cdot 3</math>, and the prime factorization of <math>384</math> is <math>2^7 \cdot 3</math>. Recalling the sum of factors formula and sum of the powers of 2 formula, we find that the sum of the factors of <math>768</math> is <math>(1+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8)(1+3^1)=(2^9-1)(4)=(511)(4)</math>, and the sum of factors of <math>384</math> is <math>1+2^1+2^2+2^3+2^4+2^5+2^6+2^7)(1+3^1)=(2^8-1)(4)=(255)(4)</math>. Then <math>f(768)=\frac{(511)(4)}{768}=\frac{511}{192}</math>, and <math>f(384)=\frac{(255)(4)}{384}=\frac{510}{192}</math>. So, <math>f(768)-f(384)=\frac{511}{192}-\frac{510}{192}=\boxed{\textbf{(B)}\ \frac{1}{192}}</math>. | ||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
− | + | == Video Solution by OmegaLearn == | |
+ | https://youtu.be/jgyyGeEKhwk?t=1116 | ||
+ | ~ pi_is_3.14 | ||
− | + | ==Video Solution by TheBeautyofMath== | |
+ | https://youtu.be/FFlj8W_xkPc | ||
− | ~ | + | ~IceMatrix |
==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}} |
Latest revision as of 13:48, 4 April 2024
Contents
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 factorizations of and
are
and
respectively. 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 factorizations of and
are
and
respectively. Note that
Therefore, the answer is
~MRENTHUSIASM
Solution 4
Note that
Both numbers take the form
. Define a function
as being the sum of the factors for any number
, where
is a nonnegative integer.
If you list out the factors of a number , all the factors are even except for
and
. So to produce a list of factors for the number
, you can multiply all the factors on the first list by
and then add on
and
. Thus,
.
We need to find . This can be written as
. Our recursive formula states that
so
. Thus, the fraction simplifies to
. Note that we never needed to figure out what
or
were, sparing us a lot of calculation.
~Curious_crow
Solution 5
We know that
We want to find the difference of these so
We are only left with the even divisors of
on the fraction on the right so the answer would be the sum of all odd divisors of
(
and
) over
:
~pengf
Solution 6
As in Solution 1, the prime factorization of is
, and the prime factorization of
is
. Recalling the sum of factors formula and sum of the powers of 2 formula, we find that the sum of the factors of
is
, and the sum of factors of
is
. Then
, and
. So,
.
~ cxsmi
Video Solution by OmegaLearn
https://youtu.be/jgyyGeEKhwk?t=1116
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
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.