SOLUTION

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 is 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.

Answer: D.
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Hi Bunuel, In statement-1, how do we know that the powers of two prime numbers is 1? I interpreted st-1 as n is a product of two different primes but their powers could be anything. So n could be ab, a^1*b, or a*b^1. Not Sufficient.
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Haha! Off course. Thats why silly errors are killing me.



Yes makes sense. thanks!
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