ziyuenlau wrote: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.
Source : e-GMAT
Official Answer : C
Hi ziyuenlau,
The smallest three prime numbers are: 2, 3, and 5
Thus, n = 2^x*3^y*5^z; where x, y, and z are the number of times 2, 3, and 5, respectively, occur in n.
We have to find out how many positive integers divide n completely.
We know that the total number of factors of n = 2^x*3^y*5^z is given by (x+1)(y+1)(z+1)
Thus, the total number of positive integers divide n completely = (x+1)(y+1)(z+1)
If we get the value of x, y and z, we get the UNIQUE answer.
Let's take each statement one by one.
S1: The total number of times the prime factors of n occur in n is 5.
=> x + y + z = 5
We cannot get the unique values of x, y and x.
Case 1: If x = y = 1, and z = 3
The total number of factors of n would be (1+1)(1+1)(3+1) = 2*2*4 =16
Case 2: If x = y = 2, and z = 1
The total number of factors of n would be (2+1)(2+1)(1+1) = 3*3*2 =18
No unique value! Not sufficient!
S2: The product of the number of times each prime factor of n occurs in n is 4.
=> xyz = 4
We cannot get the unique values of x, y and x.
Case 1: If x = y = 1, and z = 4
The total number of factors of n would be (1+1)(1+1)(4+1) = 2*2*5 = 20
Case 2: If x = y = 2, and z = 1
The total number of factors of n would be (2+1)(2+1)(1+1) = 3*3*2 = 18
No unique value! Not sufficient!
S1 and S2:
Only Case 2 works.
Thus, the total number of factors of n would be 18. Sufficient!
The correct answer:
C
Hope this helps!
Relevant book:
Manhattan Review GMAT Data Sufficiency Guide
-Jay
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