Help : Quant question from Manhattan CAT

[svshaha]

Hi Guys,

Need help in solving following question :
How many factors does 362 have?
2
8
24
25
26


The correct answer is D: 25

Can anyone please explain how come answer is 25?

Thanks in advance!!!

[tim]

Before we help with this question, we need you to show some effort of your own. What did you try on this question? Where did you get stuck?

[Rijul Negi]

Hi Guys,

Need help in solving following question :
How many factors does 362 have?
2
8
24
25
26


The correct answer is D: 25

Can anyone please explain how come answer is 25?

Thanks in advance!!!

Hi
Are you sure the no is 362? or is their something wrong in what i am doing?

When we need to find the total no of factors
what we usually do is
first convert the number into multiples of prime factors

in this case
362 = 2 * 181 [181 is a prime no]

The total no of factors would be (1+1)*(1+1) = 4
which is not in the options


What happens is
let us say we have a no z
z = a^m * b^n * c^p
here a,b and c are prime nos

when we need to select a factor.. we can choose any power of a from 0 to m
same for b=> 0 to n and for c => 0 to p

the total no of ways of selecting a factor can be rephrased as total no of factors of z => (m+1)(n+1)(p+1)

[svshaha]

Thanks guys for your messages and I must apologies for my ignorance.

It’s 36^2 (36 raise to 2) not 362.

I would solve it by normal method :
36 * 36 = 1296
1 1296
2 648
3 432
And so on..

During test, solving this questions by above method would be time consuming. So any faster way to solve this question would be helpful.

[Rijul Negi]

Thanks guys for your messages and I must apologies for my ignorance.

It’s 36^2 (36 raise to 2) not 362.

I would solve it by normal method :
36 * 36 = 1296
1 1296
2 648
3 432
And so on..

During test, solving this questions by above method would be time consuming. So any faster way to solve this question would be helpful.

I have mentioned it in my previos post...Check out my previous post..

36^2 = (3^2 x 2^2)^2 = 3^4 * 2^4

Total no of factors = (4+1)*(4+1) = 25

[svshaha]

Thanks a lot Rijul!!! That was really helpful.

[jnelson0612]

Very nice Rijul!

[clarence.booth]

I got as far as 36^2 = (3^2 x 2^2)^2 = 3^4 * 2^4

How did we make the logical leap from here to the final answer of 25?

[RonPurewal]

I got as far as 36^2 = (3^2 x 2^2)^2 = 3^4 * 2^4

How did we make the logical leap from here to the final answer of 25?

you can have up to four 3's and up to four 2's.
so, you can have
• four 3's and (four, three, two, one, or no 2's)
• three 3's and (four, three, two, one, or no 2's)
• two 3's and (four, three, two, one, or no 2's)
• one 3 and (four, three, two, one, or no 2's)
• no 3's and (four, three, two, one, or no 2's)

that's 25 cases overall. (the very last case, "no 3's and no 2's", corresponds to the factor 1.)

this same sort of thing is going to happen with any prime factorization, so it can be generalized into a formula (which the other people on this thread are presumably using).