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1990 AIME Problems/Problem 5: Difference between revisions

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== See also ==
== See also ==
{{AIME box|year=1990|num-b=4|num-a=6}}
{{AIME box|year=1990|num-b=4|num-a=6}}
== Video Solution!!! ==
https://www.youtube.com/watch?v=zlFLzuotaMU


[[Category:Intermediate Number Theory Problems]]
[[Category:Intermediate Number Theory Problems]]
{{MAA Notice}}
{{MAA Notice}}

Revision as of 08:50, 26 June 2022

Problem

Let $n^{}_{}$ be the smallest positive integer that is a multiple of $75_{}^{}$ and has exactly $75_{}^{}$ positive integral divisors, including $1_{}^{}$ and itself. Find $\frac{n}{75}$.

Solution

The prime factorization of $75 = 3^15^2 = (2+1)(4+1)(4+1)$. For $n$ to have exactly $75$ integral divisors, we need to have $n = p_1^{e_1-1}p_2^{e_2-1}\cdots$ such that $e_1e_2 \cdots = 75$. Since $75|n$, two of the prime factors must be $3$ and $5$. To minimize $n$, we can introduce a third prime factor, $2$. Also to minimize $n$, we want $5$, the greatest of all the factors, to be raised to the least power. Therefore, $n = 2^43^45^2$ and $\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}$.

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

Video Solution!!!

https://www.youtube.com/watch?v=zlFLzuotaMU These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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