Difference between revisions of "1990 AIME Problems/Problem 5"

Latest revision as of 02:45, 21 January 2023

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 $\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}$ .

Video Solution by OmegaLearn

https://youtu.be/jgyyGeEKhwk?t=588

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=zlFLzuotaMU

@@ Line 1: / Line 1: @@
 == Problem ==
-Let <math>n^{}_{}</math> be the smallest positive integer that is a multiple of <math>75_{}^{}</math> and has exactly <math>75_{}^{}</math> positive integral divisors, including <math>1_{}^{}</math> and itself. Find <math>n/75^{}_{}</math>.
+Let <math>n^{}_{}</math> be the smallest positive [[integer]] that is a multiple of <math>75_{}^{}</math> and has exactly <math>75_{}^{}</math> positive integral divisors, including <math>1_{}^{}</math> and itself. Find <math>\frac{n}{75}</math>.
 == 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>.
+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 by OmegaLearn ==
+https://youtu.be/jgyyGeEKhwk?t=588
+~ pi_is_3.14
+==Video Solution==
+https://www.youtube.com/watch?v=zlFLzuotaMU
 == See also ==
 {{AIME box|year=1990|num-b=4|num-a=6}}
+[[Category:Intermediate Number Theory Problems]]
+{{MAA Notice}}

1990 AIME (Problems • Answer Key • Resources)
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