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Let f(n) be the number of positive factors of n. What is min

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Let f(n) be the number of positive factors of n. What is min

by Max@Math Revolution » Sun Jul 07, 2019 11:26 pm

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[GMAT math practice question]

Let f(n) be the number of positive factors of n. What is minimum x satisfying f(420)*f(x)=96?

A. 4
B. 6
C. 8
D. 10
E. 12
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by swerve » Tue Jul 09, 2019 5:56 pm
Given, \(f(420) =\) number of positive factors of 420
\(\Rightarrow 420 = 2^2\cdot 3\cdot 5 \cdot 7\)

Number is positive factors \(= (2 + 1)(1 + 1)(1 + 1)(1 + 1) = 24\)

\(f(420)\cdot f(x) = 96\)
\(\Rightarrow 24\cdot f(x) = 96\)
\(\Rightarrow f(x) = 4\)

So, the number of positive factors of \(x = 4\)

Number of positive factors of
A. \(4 = 2^2 = 2 + 1 = 3 \Rightarrow \) Wrong
B. \(6 = 2\cdot 3 = (1 + 1)(1 + 1) = 4 \Rightarrow\) Satisfies
C. \(8 = 2^3 = 3 + 1 = 4 \Rightarrow \) Satisfies
D. \(10 = 2\cdot 5 = (1 + 1)(1 + 1) = 4 \Rightarrow \) Satisfies
E. \(12 = 2^2\cdot 3 = (2 + 1)(1 + 1) = 6\Rightarrow \) Wrong

So, the minimum value that satisfies \(= 6\,\,{\color{green}\checkmark}\)
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by Max@Math Revolution » Tue Jul 09, 2019 11:19 pm
=>

Since 420 = 2^2*3*5*7, we have f(420) = (2+1)(1+1)(1+1)(1+1) = 24 and f(x) = 96 / f(420) = 96 / 24 = 4.
The smallest integer with 4 factors is 6, which has 4 factors, 1, 2, 3 and 6.

Therefore, B is the answer.
Answer: B
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