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Help needed

by Soumita Ghosh » Fri Jun 22, 2012 1:10 am
How many different positive integers are factors of 1089?

A) 11

B) 7

C) 4

D) 9

E) 6

My answer is D .

I have found the factors by this method :

1089/3 = 363

363/3 =121

121/11 =11

11/11 =1

Now 3x3x11x11 are the prime factors of 1089. From these prime factors I can find all the possible factors of 1089. To do so I have to think of every possible unique combination that I can come up with using 3,3,11 and 11

3
11
3x3
11x11
3x11
3x3x11
11x11x3

are the sever possible unique combinations that I can come up with.Further 1 and 1089 are there too. So total of 9 factors.

But the answer OA is C . Please explain where I am wrong .
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by gmat_and_me » Fri Jun 22, 2012 4:33 am
If you have posted the question right, then D is the answer. In
general, if you break down a number into it's prime factors,

X = (p)^a * (p1)^b * ...

then the number of different factors will be (a+1) * (b+1) *...

and so on...

HTH
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by Anurag@Gurome » Fri Jun 22, 2012 6:10 am
Soumita Ghosh wrote:How many different positive integers are factors of 1089?
...
My answer is D
What you have is correct.
But there is a method to solve this kind of problem rather than calculating the number of factors manually.

Number of different positive factors of an integer whose prime factorization is of the form (p^a)*(q^b)*(r^c)... is (a + 1)(b + 1)(c + 1)...

In this case, 1089 = (3^2)*(11^2)
Hence, number of different positive factors of 1089 = (2 + 1)*(2 + 1) = 3*3 = 9

The correct answer is D.
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by Soumita Ghosh » Fri Jun 22, 2012 3:52 pm
Thanks a lot Anurag . I will follow the method you mentioned above from next time.
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by Soumita Ghosh » Fri Jun 22, 2012 3:52 pm
Thanks a lot Anurag ..I will follow your method from next time..
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