Difference between revisions of "1990 AIME Problems/Problem 5"
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The [[prime factorization]] of <math>75 = 3^15^2 = (2+1)(4+1)(4+1)</math>. For <math>n</math> to have exactly <math>75</math> integral divisors, we need to have <math>n = p_1^{e_1-1}p_2^{e_2-1}\cdots</math> such that <math>e_1e_2 \cdots = 75</math>. Since <math>75|n</math>, two of the [[prime]] [[factor]]s must be <math>3</math> and <math>5</math>. To minimize <math>n</math>, we can introduce a third prime factor, <math>2</math>. Also to minimize <math>n</math>, we want <math>5</math>, the greatest of all the factors, 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}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}</math>. | The [[prime factorization]] of <math>75 = 3^15^2 = (2+1)(4+1)(4+1)</math>. For <math>n</math> to have exactly <math>75</math> integral divisors, we need to have <math>n = p_1^{e_1-1}p_2^{e_2-1}\cdots</math> such that <math>e_1e_2 \cdots = 75</math>. Since <math>75|n</math>, two of the [[prime]] [[factor]]s must be <math>3</math> and <math>5</math>. To minimize <math>n</math>, we can introduce a third prime factor, <math>2</math>. Also to minimize <math>n</math>, we want <math>5</math>, the greatest of all the factors, 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}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}</math>. | ||
| − | ==Video Solution | + | ==Video Solution== |
https://www.youtube.com/watch?v=zlFLzuotaMU | https://www.youtube.com/watch?v=zlFLzuotaMU | ||
Revision as of 12:16, 26 June 2022
Contents
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 
. For 
to have exactly 
integral divisors, we need to have 
such that 
. Since 
, two of the prime factors must be 
and 
. To minimize , we can introduce a third prime factor, 
. Also to minimize , we want , the greatest of all the factors, to be raised to the least power. Therefore, 
and 
.
Video Solution
https://www.youtube.com/watch?v=zlFLzuotaMU
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
