Difference between revisions of "2021 AMC 12B Problems/Problem 7"

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<math>\boxed{\textbf{(C)} ~1 : 14}</math>
 
<math>\boxed{\textbf{(C)} ~1 : 14}</math>
  
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)\rightarrow\boxed{\textbf{(C)}}</math>
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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)\rightarrow\boxed{\textbf{(C)}}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2021|ab=B|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:08, 11 February 2021

Problem

Let $N = 34 \cdot 34 \cdot 63 \cdot 270$. What is the ratio of the sum of the odd divisors of $N$ to the sum of the even divisors of $N$?

$\textbf{(A)} ~1 : 16 \qquad\textbf{(B)} ~1 : 15 \qquad\textbf{(C)} ~1 : 14 \qquad\textbf{(D)} ~1 : 8 \qquad\textbf{(E)} ~1 : 3$

Solution

$\boxed{\textbf{(C)} ~1 : 14}$

Prime factorize $N$ to get $N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}$. For each odd divisor $n$ of $N$, there exist even divisors $2n, 4n, 8n$ of $N$, therefore the ratio is $1:(2+4+8)\rightarrow\boxed{\textbf{(C)}}$

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
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

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