Difference between revisions of "2021 AMC 12B Problems/Problem 7"
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+ | {{duplicate|[[2021 AMC 10B Problems#Problem 12|2021 AMC 10B #12]] and [[2021 AMC 12B Problems#Problem 7|2021 AMC 12B #7]]}} | ||
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==Problem== | ==Problem== | ||
Let <math>N = 34 \cdot 34 \cdot 63 \cdot 270</math>. What is the ratio of the sum of the odd divisors of <math>N</math> to the sum of the even divisors of <math>N</math>? | Let <math>N = 34 \cdot 34 \cdot 63 \cdot 270</math>. What is the ratio of the sum of the odd divisors of <math>N</math> to the sum of the even divisors of <math>N</math>? | ||
<math>\textbf{(A)} ~1 : 16 \qquad\textbf{(B)} ~1 : 15 \qquad\textbf{(C)} ~1 : 14 \qquad\textbf{(D)} ~1 : 8 \qquad\textbf{(E)} ~1 : 3</math> | <math>\textbf{(A)} ~1 : 16 \qquad\textbf{(B)} ~1 : 15 \qquad\textbf{(C)} ~1 : 14 \qquad\textbf{(D)} ~1 : 8 \qquad\textbf{(E)} ~1 : 3</math> | ||
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− | Prime factorize <math>N</math> to get <math>N=2^{3}3^{5}5\cdot 7\cdot 17^{2}</math>. For each odd divisor <math>n</math> of <math>N</math>, there exist even divisors <math>2n, 4n, 8n</math> of <math>N</math>, therefore the ratio is <math>1:(2+4+8)\ | + | ==Solution 1== |
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+ | Prime factorize <math>N</math> to get <math>N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}</math>. For each odd divisor <math>n</math> of <math>N</math>, there exist even divisors <math>2n, 4n, 8n</math> of <math>N</math>, therefore the ratio is <math>1:(2+4+8)=\boxed{\textbf{(C)} ~1 : 14}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Prime factorizing <math>N</math>, we see <math>N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}</math>. The sum of <math>N</math>'s odd divisors are the sum of the factors of <math>N</math> without <math>2</math>, and the sum of the even divisors is the sum of the odds subtracted by the total sum of divisors. The sum of odd divisors is given by <cmath>a = (1+3+3^2 + 3^3 + 3^4 + 3^5)(1 + 5)(1+7)(1+17+17^2)</cmath> and the total sum of divisors is <cmath>(1+2+4+8)(1+3+3^2 + 3^3 + 3^4 + 3^5)(1 + 5)(1+7)(1+17+17^2) = 15a.</cmath> Thus, our ratio is <cmath>\frac{a}{15a-a} = \frac{a}{14a} = \boxed{\textbf{(C)} ~1 : 14}.</cmath> | ||
+ | |||
+ | ~JustinLee2017 | ||
+ | |||
+ | ==Video Solution by Punxsutawney Phil== | ||
+ | https://youtube.com/watch?v=qpvS2PVkI8A&t=643s | ||
+ | |||
+ | == Video Solution by OmegaLearn (Prime Factorization) == | ||
+ | https://youtu.be/U3msAYWeMbI | ||
+ | |||
+ | ~ 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== | ||
+ | https://youtu.be/duZG-jirKRc | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2021|ab=B|num-b=6|num-a=8}} | ||
+ | {{AMC10 box|year=2021|ab=B|num-b=11|num-a=13}} | ||
+ | {{MAA Notice}} |
Latest revision as of 06:49, 12 June 2021
- 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. The sum of odd divisors is given by and the total sum of divisors is Thus, our ratio is
~JustinLee2017
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.