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factor question

Expert replies
Source: — Problem Solving |

by sudhir3127 » Tue Aug 05, 2008 8:11 am
30= 2 *3*5

hence its (1+1)*(1+1)*(1+1) number of factors..

thus eight...
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Re: factor question

by manju_ej » Tue Aug 05, 2008 8:21 am
30 = 2*3*5

So factors are 1,2,3,5,6,10,15,30.

Apart from 1,2,3,5, and 30, find the product of 2,3, and 5 in all combinations.
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by pepeprepa » Tue Aug 05, 2008 9:06 am
Thanks for your formula Sudhir, it helps a lot.
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by vaivish » Sat Aug 09, 2008 6:54 am
thanks....
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by 2008 » Sat Aug 09, 2008 8:38 am
sudhir3127 wrote:30= 2 *3*5

hence its (1+1)*(1+1)*(1+1) number of factors..

thus eight...

maybe i m just burn out this afternoon, but i dont get it... can you please explain this formula?
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by Ian Stewart » Sat Aug 09, 2008 9:41 am
2008 wrote:
sudhir3127 wrote:30= 2 *3*5

hence its (1+1)*(1+1)*(1+1) number of factors..

thus eight...

maybe i m just burn out this afternoon, but i dont get it... can you please explain this formula?
If you need to count the number of divisors of x:

1) prime factorize x, and write the prime factorization in the conventional notation (using exponents if primes are repeated);
2) now look only at the exponents: add one to each exponent, and multiply the resulting numbers.

To take an example, how many positive divisors does 180 have?

1) Prime factorize: 180 = (2^2)(3^2)(5^1)
2) Add one to each power and multiply: 3*3*2 = 18.

So 180 has 18 different positive divisors. You can see why this works: any number that can be written as (2^a)(3^b)(5^c) will be a divisor of (2^2)(3^2)(5^1) just as long as:

a = 0, 1 or 2 (three choices)
b = 0, 1 or 2 (three choices)
c = 0 or 1 (two choices)

Now it's a counting problem- we multiply the number of choices we have for each of a, b and c: 3*3*2 = 18.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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by 2008 » Sat Aug 09, 2008 11:52 am
that's genius :D but 2 weeks before the exam it cab be a bit scaring... thanks!
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