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If n is an integer, then n divisible by how many positive integers? (1) n is the product of two different prime numbers. (2) n and 2^3 are each divisible by the same number of positive integers.

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. For more on number properties check: math-number-theory-88376.html

BACK TO THE ORIGINAL QUESTION:

If n is an integer, then n divisible by how many positive integers?

(1) n is the product of two different prime numbers --> n=ab, where a and b are primes, so # of factors is (1+1)(1+1)=4. Sufficient.

(2) n and 2^3 are each divisible by the same number of positive integers --> 2^3 has 4 different positive factors (1, 2, 4, and 8) so n has also 4. Sufficient.

It took me just 15 seconds to solve this.. N is a product of 2 different prime nos.......then 1,n and dose two prime nos. are divisible by n ...hence 4 nos. agen, 2^3 = 8, has 4 nos. from which it can be divided...agen n is divisible by 4 nos. Hence, D

Re: If n is an integer, then n divisible by how many positive [#permalink]
31 Jan 2013, 12:10

@ Fozzzy Statement 1 - n is the product of two different integers . They may be 2*3 or 3*7 or any two integers. Since they yield different products. We cannot determine the # of factors for n. Hence Statement 1 - Insufficient. Statement 2 - n and 2^3 are each divisible by the same number of positive integers. 2^3 = 8. Having 4 factors (1,2,4,8) . Since the statement says n and 8 are divisible by the same num of integers. n=4. Hence Statement 2 - Sufficient

Re: If n is an integer, then n divisible by how many positive [#permalink]
26 Aug 2014, 19:29

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