2021 AMC 12B Problems/Problem 7
- The following problem is from both the 2021 AMC 10B #12 and 2021 AMC 12B #7, so both problems redirect to this page.
Contents
Problem
Let . What is the ratio of the sum of the odd divisors of
to the sum of the even divisors of
?
Solution 1
Prime factorize to get
. For each odd divisor
of
, there exist even divisors
of
, therefore the ratio is
Solution 2
Prime factorizing , we see
.
The sum of
's odd divisors are the sum of the factors of
without
, and the sum of the even divisors is the sum of the odds subtracted by the total sum of divisors.
BY SUM OF FACTORS FORMULA (search if you don' know):
The formula is actually for all factors, but we can just take out the , so we have:
The sum of odd divisors is given by and the total sum of divisors is
Thus, our ratio is
~JustinLee2017 ~minor edits by SwordAxe
Solution 3
Prime factorizing , we have that there is
in our factorization. Now, call the sum of the odd divisors
. We know that if we multiply k by 2, we will have even divisors. So, we can multiply k by 2,
respectively to get 14k as the sum of the even divisors. Therefore, the answer is
~MC
Video Solution (Under 2 min!)
~Education, the Study of Everything
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=qpvS2PVkI8A&t=643s
Video Solution by OmegaLearn (Prime Factorization)
~ pi_is_3.14
Video Solution by Hawk Math
https://www.youtube.com/watch?v=VzwxbsuSQ80
Video Solution by TheBeautyofMath
https://youtu.be/L1iW94Ue3eI?t=478
~IceMatrix
Video Solution by Interstigation
~Interstigation
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 |
2021 AMC 10B (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 10 Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.