If the number of factors of a positive integer,

Question

If the number of factors of a positive integer, \(P\), that are in the form \(5k\) is \(30\), and \(P = 5^a * P_1^b * P_2^c\), where \(5\), \(P_1\) & \(P_2\) are distinct prime factors and a, b, & c are positive integers, then what is the value of a + b + c?

(1) The total number of factors of \(P\) is even.

(2) P is divisible by 125, but not divisible by 625.

(1) The total number of factors of  P  is even.

(2) P is divisible by 125, but not divisible by 625.

Nidhijain94 · Answer

Not well formed question. Please check. I dont need any options to answer. From question stem itself we can say a+b+c will be 7

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gmatophobia · Answer

Nidhijain94  - Can you please share some details on your solution? I am unable to follow how did we conclude that " a+b+c will be 7 " from the question stem without using any options.

I don't see any issue with the question, hence wanted to understand your solution.

avirupm97 · Answer

This is a completely valid and good question, nothing wrong here. And no, a + b + c is neither 7 nor is it deducible from the question stem.

Let's start with the number N = 5^a ∗ P1 ^ b ∗ P2 ^ c. It is given that factors that are multiple of 5=30.

Statement A: Total factors of the number is even. So, factors of 5 + Non 5 factors = even. We already know that factors of 5 is even(30), but no idea about b and c. So, insufficient.

Statement 2: P is divisible by 125, but not divisible by 625. So, the highest power of 5 is 5^3, not 5^4. This is good news. Gives an idea about power a.
Good thing that question asks for a+b+c and not individual a, b and c values. Let's see if we can get b+c somehow.

Total factors that this number can have: (3+1)(b+1)(c+1) => 4(b+1)(c+1)
Out of these factors, Total factors which are in the form 5k(multiple of 5) = 30
Remaining factors which are not a multiple of 5 = (b+1)(c+1)

So total factors = # of factors which are a multiple of 5 + # factors which are not a multiple of 5, so equating, 
=> 4(b+1)(c+1) = (b+1)(c+1) + 30
=> 3(b+1)(c+1) = 30
=> (b+1)(c+1) = 10

(b+1)(c+1) can give product 10 if it is in the form of 5*2 or 2*5 (10*1 or 1*10 is not possible since a, b and c are positive integers, 1 will equate to either b or c being 0)
So, b can be either 4 or 1, and correspondingly c can be either 1 or 4, wither way b+c=5.

We already know that a=3(125 = 5^3), so a+b+c = 3+5 = 8

We can eliminate option C, D and E. 
Answer B.

aditya khanna · Answer

This question is flawed since given that number of factors of a positive integer, P , that are in the form 5k is 30
it can be written as a(b+1)(c+1) = 30
30 can be factorize as 2x3x5 
and we want to know the value of a+b+c = a+(b+1)+(c+1) -2 = 10-2 
=8 
Answer is achieived without using any condition .

If the number of factors of a positive integer,

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