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Difference between revisions of "2005 AMC 12B Problems/Problem 21"
(→‎Solution: original solution wasn't quite correct since the prime factorization might not be p^1q^1r^2s^4. Improved solution using more standard notation and multiplicativity of d(n).)
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== Solution ==
 
== Solution ==
−
If <math>n</math> has <math>60</math> factors, then <math>n</math> is a product of <math>2\times2\times3\times5</math> powers of (not necessarily distinct) primes. When multiplied by <math>7</math>, the amount of factors of <math>n</math> increased by <math>\frac{80}{60}=\frac{4}{3}</math>, so there are <math>4</math> possible powers of <math>7</math> in the factorization of <math>7n</math>, and <math>3</math> possible powers of <math>7</math> in the factorization of <math>n</math>, which would be <math>7^0</math>, <math>7^1</math>, and <math>7^2</math>. Therefore the highest power of <math>7</math> that could divide <math>n</math> is <math>2\Rightarrow\boxed{C}</math>.
+
We may let <math>n = 7^k \times m</math>, where <math>m</math> is not divisible by 7. Using the fact that the number of divisors function <math>d(n)</math> is multiplicative, we have <math>d(n) = d(7^k)d(m) = (k+1)d(m) = 60</math>. Also, <math>d(7n) = d(7^{k+1})d(m) = (k+2)d(m) = 80</math>. These numbers are in the ratio 3:4, so <math>\frac{k+1}{k+2} = \frac{3}{4} \implies k = 2 \Rightarrow \boxed{C}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2005|ab=B|num-b=20|num-a=22}}
 
{{AMC12 box|year=2005|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:43, 23 May 2018

Problem

A positive integer $n$ has $60$ divisors and $7n$ has $80$ divisors. What is the greatest integer $k$ such that $7^k$ divides $n$?

$\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}$

Solution

We may let $n = 7^k \times m$, where $m$ is not divisible by 7. Using the fact that the number of divisors function $d(n)$ is multiplicative, we have $d(n) = d(7^k)d(m) = (k+1)d(m) = 60$. Also, $d(7n) = d(7^{k+1})d(m) = (k+2)d(m) = 80$. These numbers are in the ratio 3:4, so $\frac{k+1}{k+2} = \frac{3}{4} \implies k = 2 \Rightarrow \boxed{C}$.

See also

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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