Difference between revisions of "2005 AMC 12B Problems/Problem 21"
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== 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>? | ||
| + | |||
| + | <math>\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}</math> | ||
== Solution == | == Solution == | ||
| + | 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 == | ||
| − | + | {{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 
has 
divisors and 
has 
divisors. What is the greatest integer 
such that 
divides ?
Solution
We may let 
, where 
is not divisible by 7. Using the fact that the number of divisors function 
is multiplicative, we have 
. Also, 
. These numbers are in the ratio 3:4, so 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
