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A perfect sq. has three positive divisors..

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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A perfect sq. has three positive divisors..

by uptowngirl92 » Wed Aug 19, 2009 5:42 pm
I was wondering if anyone can help me with the explanation for question 241 in the OG11.

If the integer n has exactly three positive divisors, including 1 and n how many positive divisors does n^2 have?

I dont understand the concept that if a no. has 3 positive divisors it is a perfect sq..HOW???WHY??
We know..
most nos. have even no. od divisors.
prime nos. have exactly 2 divisors,1 and itself..

Can someone follow the same reasonung and explain the concept to me?
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by tohellandback » Wed Aug 19, 2009 8:22 pm
if a number has exactly 3 positive factors:
n=p^2, where p is a prime number
and thats why n must be the square of a prime number

n^2=p^4
so n^2 will have 5 positive factors.
The powers of two are bloody impolite!!
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by Michael Birdsall » Sun Sep 06, 2009 3:54 am
There is a relationship between the powers of the primes that make up an integer and the number of positive factors.

The number of positive factors will equal to the product of powers of each of the primes+1.

For example, suppose that p, q, and r are prime numbers then (p^2)(q^3)(r^1) will have (2+1)(3+1)(1+1)=24 positive factors, because 2, 3, and 1 are the powers of p, q, and r respectively.

In the question we are told that integer n has three positive factors, so we can apply the rule above to it.

3 must equal the product of the powers of primes+1 in n. To get the product 3 we must have multiplied together (3 and 1). So the powers of the primes in n must have been (3-1)=2 and (1-1)=0. Since any base raised to the zero power must be 1, then we need not consider a prime raised to the zero power. So, we are left with one prime which must have been raised to the power of 2.

The rule above is pretty important as far as being able to figure out the number of positive divisors (factors) of an integer. I have seen it explicitly tested on the GMAT.
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