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Arithmetic

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Source: — Problem Solving |
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by cramya » Thu Mar 12, 2009 2:49 pm
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
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by franciskyle » Thu Mar 12, 2009 2:52 pm
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.
Last edited by franciskyle on Thu Mar 12, 2009 2:55 pm, edited 1 time in total.
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by Alara533 » Thu Mar 12, 2009 2:53 pm
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
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by lolo » Thu Mar 12, 2009 3:04 pm
Thank you guys for these great explanations.

Do you have other ways to find the OA : 8 ?
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by franciskyle » Thu Mar 12, 2009 7:22 pm
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 :)
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by cramya » Thu Mar 12, 2009 8:00 pm
Frankie,
Giving credit to the source:

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

Hope this helps!

Regards,
CR
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