Arithmetic

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lolo: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Sat Dec 13, 2008 9:00 am

Arithmetic

by lolo » Thu Mar 12, 2009 2:44 pm

For how many positive integers x is 130/x an integer ?

8
7
6
5
3

Could somebody please help me solving this intermediate question ?

I need some explanations as I always find the wrong value.

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Source: Beat The GMAT — Problem Solving | Thu Mar 12, 2009 2:44 pm

cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pm; Thanked: 331 times; Followed by:11 members

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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franciskyle: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Wed Mar 11, 2009 9:56 pm; Location: Whitler, BC, Canada

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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Alara533: Senior | Next Rank: 100 Posts; Posts: 73; Joined: Wed Jan 28, 2009 9:00 am; Thanked: 6 times; GMAT Score:680

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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lolo: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Sat Dec 13, 2008 9:00 am

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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franciskyle: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Wed Mar 11, 2009 9:56 pm; Location: Whitler, BC, Canada

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