Difference between revisions of "2005 AMC 12B Problems/Problem 21"

(Solution)
(Solution)
 
(6 intermediate revisions by 5 users not shown)
Line 1: Line 1:
{{empty}}
 
 
== Problem ==
 
== Problem ==
 
A positive integer <math>n</math> has <math>60</math> divisors and <math>7n</math> has <math>80</math> divisors.  What is the greatest integer <math>k</math> such that <math>7^k</math> divides <math>n</math>?
 
A positive integer <math>n</math> has <math>60</math> divisors and <math>7n</math> has <math>80</math> divisors.  What is the greatest integer <math>k</math> such that <math>7^k</math> divides <math>n</math>?
Line 6: Line 5:
  
 
== 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 before there were <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 \cdot 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{\mathrm{C}}</math>.
  
 
== See also ==
 
== See also ==
* [[2005 AMC 12B Problems]]
+
{{AMC12 box|year=2005|ab=B|num-b=20|num-a=22}}
 +
{{MAA Notice}}

Latest revision as of 19:04, 24 December 2020

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 $n = 7^k \cdot m$, 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 $\frac{k+1}{k+2} = \frac{3}{4} \implies k = 2 \Rightarrow \boxed{\mathrm{C}}$.

See also

2005 AMC 12B (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png