Arithmetic

Basically its asking how many factors does 130 have?

The easy way to do it would be to break down 130 in to its prime factors first

130 =2*5*13 = 2^1*5^1*13^1


Now to find the number of factors add 1 to the exponent of each prime factor and multiply

(1+1) (1+1) (1+1) = 8 unique factors


Illustarting with another eg:

For 12 = 2*3*2 = 2^2 * 3^1

No of factors would be (2+1) (1+1) = 6

12 has 6 unique factors


Hope this helps!

Regards,
CR

To do this quicker, you could have looked at the numerator. Since it ends in 0, you automatically know that it is divisible by 1,2,5,10. That gives us 4 factors, which need each some other unique factor to multiply with to get 130 - therefore you immediately know that there are AT LEAST 8 factors. Since 8 is the highest option, you can quickly select it.

Is the answer 8?

Approach-

For 130/x to be an integer, x must be a factor of 130

Find prime factors of 130 -> 2,5,13 --> 3 factors
The multiples of these factors will also be factors of 130, ie
5 x 2 = 10
5 x 13 = 65
2 x 13 = 26
5 x 2 x 13 = 130

and finally 1 is also a factor of 130...so total 8 possible values for x

Thank you guys for these great explanations.

Do you have other ways to find the OA : 8 ?

Hi cramya...

I find the way you solved this problem interesting.

cramya wrote:
The easy way to do it would be to break down 130 in to its prime factors first

130 =2*5*13 = 2^1*5^1*13^1


Now to find the number of factors add 1 to the exponent of each prime factor and multiply

(1+1) (1+1) (1+1) = 8 unique factors

Are you able to explain the theory behind your answer? Why should I expect that when I add 1 to each exponent of the prime factors and then multiply them I will arrive at the total number of factors?

I'm sorry if this is silly, but I am having trouble piecing this logic.

Thanks

Frankie,
Giving credit to the source:

https://www.gmathacks.com/gmat-math/numb ... teger.html

Hope this helps!

Regards,
CR