30= 2 *3*5
hence its (1+1)*(1+1)*(1+1) number of factors..
thus eight...
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factor question
Source: Beat The GMAT — Problem Solving |
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.
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.
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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If you need to count the number of divisors of x: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?
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
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but 2 weeks before the exam it cab be a bit scaring... thanks!
