If one of the factors of the number $3^{7} 2^{8} 17^{3}$ is randomly chosen, then the probability that the chosen factor will be a perfect square is

Question and options

If one of the factors of the number

3^{7} 2^{8} 17^{3}

is randomly chosen, then the probability that the chosen factor will be a perfect square is

$\frac{5}{36}$
$\frac{1}{12}$
$\frac{3}{40}$
$\frac{5}{32}$

Answer and explanation

Correct answer: A - $\frac{5}{36}$

Number of factors =8*9*4=288. Square factors: 3-exponent in {0,2,4,6} (4), 2-exponent in {0,..,8 even} (5), 17-exponent {0,2} (2): 40. Probability 40/288=5/36.

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If one of the factors of the number \(3^{7} 2^{8} 17^{3}\) is randomly chosen, then the probability that the chosen factor will be a perfect square is