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Re: M01-Q35 Please help with this question. thanks : Retired Discussions [Locked]

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Re: M01-Q35 Please help with this question. thanks [#permalink]
04 Apr 2009, 20:09

4

This post received KUDOS

Prime factorize 90 90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

Re: M01-Q35 Please help with this question. thanks [#permalink]
30 Jan 2010, 12:43

cramya wrote:

Prime factorize 90 90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12

I don't have 12 in my answer choices and i answered incorrectly Although i understood the question perfectly

If @x is the number of distinct positive divisors of x, what is the value of @@90 ?

Re: M01-Q35 Please help with this question. thanks [#permalink]
22 Sep 2010, 05:13

cramya wrote:

Prime factorize 90 90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12

thanks:)

but could you please explain why addition of each of the exponents can produce the result? I cannot understand this:(

Re: M01-Q35 Please help with this question. thanks [#permalink]
22 Sep 2010, 05:18

cnrnld wrote:

cramya wrote:

Prime factorize 90 90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12

thanks:)

but could you please explain why addition of each of the exponents can produce the result? I cannot understand this:(

Re: M01-Q35 Please help with this question. thanks [#permalink]
22 Sep 2010, 07:43

shrouded1 wrote:

cnrnld wrote:

cramya wrote:

Prime factorize 90 90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12

thanks:)

but could you please explain why addition of each of the exponents can produce the result? I cannot understand this:(

Re: M01-Q35 Please help with this question. thanks [#permalink]
22 Sep 2010, 07:57

4

This post received KUDOS

Expert's post

cnrnld wrote:

thanks:)

but could you please explain why addition of each of the exponents can produce the result? I cannot understand this:(

MUST KNOW FOR GMAT:

Finding the Number of Factors of an Integer

First make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers.

The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2

Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors.

Back to the original question:

If @x is the number of distinct positive divisors of x , what is the value of @@90? (A) 3 (B) 4 (C) 5 (D) 6 (E) 7

The question defines @x as the number of distinct positive divisors of x. Say @6=4, as 6 have 4 distinct positive divisors: 1, 2, 3, 6.

Question: @@90=?

90=2*3^2*5, which means that the number of factors of 90 is: (1+1)(2+1)(1+1)=12. So @90=12 --> @@90=@12 --> 12=2^2*3, so the number of factors of 12 is: (2+1)(1+1)=6.

Re: M01-Q35 Please help with this question. thanks [#permalink]
26 Sep 2011, 08:45

To get number of positive factors of any number follow below 4 steps:

1. The prime factorization raised to the power: 90= (3^2)(2^1)(5^1)(1^0) 2. The powers: 2, 1, 1, 0 3. Add one to the powers: 3,2,2,1 4. Multiply the results: (3)(2)(2)(1)=12

and repeat the steps for 12. thats it you are done! _________________

Re: M01-Q35 Please help with this question. thanks [#permalink]
26 Sep 2012, 05:57

Use the Fundamental Counting Theory...

Prime factorization of 90= 2*3^2*5 Add 1 to each exponent and multiply= (2)(3)(2)=12 distinct positive divisors of 90 Prime factorizatino of 12=3*2^2 (2)(3)=6 distinct factors of 12

Answer is D

gmatclubot

Re: M01-Q35 Please help with this question. thanks
[#permalink]
26 Sep 2012, 05:57