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Factors!

by Ozlemg » Fri Jul 15, 2011 3:09 am
How many distinct integers are factors of 90?

(A) 6
(B) 8
(C) 9
(D) 10
(E) 12

I know how to solve it; but want to learn whether there is an quick way to calculate if the number is high, such as 7654.

OA : E
Tnx
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by Anurag@Gurome » Fri Jul 15, 2011 3:48 am
Ozlemg wrote:How many distinct integers are factors of 90?...

I know how to solve it; but want to learn whether there is an quick way to calculate if the number is high, such as 7654
To find the number of distinct factors of an integer N, we have to first express the integer as the product of prime integers like N = (p^a)*(q^b)...

Then the number of distinct factors of integer N is (a + 1)(b + 1)...

Here, 90 = (2^1)*(3^2)*(5^1)
Hence, number of distinct integers of 90 is (1 + 1)*(2 + 1)*(1 + 1) = 2*3*2 = 12

If the number is 7654, then 7654 = 2*43*89
Hence, number of distinct integers of 7654 is (1 + 1)*(1 + 1)*(1 + 1) = 2*2*2 = 8
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by pemdas » Fri Jul 15, 2011 3:49 am
your posted q. ==> perform prime factorization of 90
90(2)45
45(3)15
15(3)5
5(5)1 ==> (2^1)*(3^2)*(5^1), the number of distinct factors will be (1+1)*(2+1)*(1+1) by adding 1 to the power of each prime factor, there will be 12 distinct factors

the same solution way should be applied to 7654

7654(2)3827
3827(43)89
89(89)1

(2^1)*(43^1)*(89^1) ==> (1+1)*(1+1)*(1+1)=8, so 8 distinct factors

use this link for prime factorization of big numbers https://www.math.wustl.edu/cgi-bin/primes
Ozlemg wrote:How many distinct integers are factors of 90?

(A) 6
(B) 8
(C) 9
(D) 10
(E) 12

I know how to solve it; but want to learn whether there is an quick way to calculate if the number is high, such as 7654.

OA : E
Tnx
Success doesn't come overnight!
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