A positive integer n has the smallest 3 prime numbers as its

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A positive integer n has the smallest 3 prime numbers as its only prime factors. How many positive integers divide n completely?

1) The total number of times the prime factors of n occur in n is 5.
2) The product of the number of times each prime factor of n occurs in n is 4.

OA C

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by Jay@ManhattanReview » Sun Mar 24, 2019 8:37 pm

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AAPL wrote:e-GMAT

A positive integer n has the smallest 3 prime numbers as its only prime factors. How many positive integers divide n completely?

1) The total number of times the prime factors of n occur in n is 5.
2) The product of the number of times each prime factor of n occurs in n is 4.

OA C
The smallest 3 prime numbers are 2, 3 and 5.

Say n = 2^a * 3^b * 5^c, where a, b and c are positive integers

The number of factors of n are (a + 1) * (b + 1) * (c + 1). We have to get the value of (a + 1) * (b + 1) * (c + 1).

Let's take each statement one by one.

1) The total number of times the prime factors of n occur in n is 5.

=> a + b + c = 5

Case 1: Say a = 1, b = 2 and c = 2; then the number of factors of n = (a + 1) * (b + 1) * (c + 1) = (1 + 1) * (2 + 1) * (2 + 1) = 2*3*3 = 18.
Case 2: Say a = 1, b = 1 and c = 3; then the number of factors of n = (a + 1) * (b + 1) * (c + 1) = (1 + 1) * (1 + 1) * (3 + 1) = 2*2*4 = 16.

No unique value. Insufficient.

2) The product of the number of times each prime factor of n occurs in n is 4.

=> a*b*c = 4

Case 1: Say a = 1, b = 2 and c = 2; then the number of factors of n = (a + 1) * (b + 1) * (c + 1) = (1 + 1) * (2 + 1) * (2 + 1) = 2*3*3 = 18.
Case 2: Say a = 1, b = 1 and c = 4; then the number of factors of n = (a + 1) * (b + 1) * (c + 1) = (1 + 1) * (1 + 1) * (4 + 1) = 2*2*5 = 20.

No unique value. Insufficient.

(1) and (2) together

The only common value in the two statements is 18. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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