Difference between revisions of "1990 AIME Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | {{ | + | The [[prime factorization]] of <math>75 = 3^15^2</math>. Thus, for <math>n</math> to have exactly <math>75</math> integral divisors, we need to have <math>n = a^{(5 - 1)}b^{(5 - 1)}c^{(3 - 1)}</math>. Since we know that <math>n</math> is divisible by <math>75</math>, two of the factors must be <math>3</math> and <math>5</math>. To minimize <math>n</math>, the last factor must be <math>2</math>; also to minimize it, we want <math>5<math> to be raised to the least power. Therefore, <math>n = 2^43^45^2</math> and <math>\frac{n}{75} = \frac{2^43^45^2}{35^2} = 16 \cdot 27 = 432</math>. |
== See also == | == See also == | ||
{{AIME box|year=1990|num-b=4|num-a=6}} | {{AIME box|year=1990|num-b=4|num-a=6}} | ||
Revision as of 19:53, 2 March 2007
Problem
Let 
be the smallest positive integer that is a multiple of 
and has exactly positive integral divisors, including 
and itself. Find 
.
Solution
The prime factorization of 
. Thus, for 
to have exactly 
integral divisors, we need to have 
. Since we know that is divisible by , two of the factors must be 
and 
. To minimize , the last factor must be 
; also to minimize it, we want 
and 
.
See also
| 1990 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
