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OG 11 PS question

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Re: OG 11 PS question

by ssmiles08 » Sat Jun 27, 2009 4:29 am
varunkh70 wrote:If the integer n has exactly three positive divisors, including 1 and n, how many positive divisors does n^2 have?

OA later. any shortcuts?

~Varun
IMO 5

i took n as 4.

4^2 = 16

16 = 2^4

(4+1) = 5 Total factors
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by varunkh70 » Sat Jun 27, 2009 5:05 am
thanks but how did that 4+1 come ther?
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by ssmiles08 » Sat Jun 27, 2009 5:24 am
varunkh70 wrote:thanks but how did that 4+1 come ther?
anytime you have a number, you want to find its total factors, you first break it down into primes, ex. 12 = 2^2 * 3^1

Then, you add 1 to their exponents and multiply each other. Here in this case for 12, it would be (2+1)*(1+1) = 3*2 = 6 total factors.

Ian has discussed it in detail here:

https://www.beatthegmat.com/what-s-your- ... 64-15.html
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by shanmugam.d » Sat Jun 27, 2009 6:42 am
Yeah it should be 5
- a number can have exactly 3 factor if it is a square of a prime number
- hence square of a number which is a prime square will be in the form:

- say x is a prime no., n = x^2
- n^2 =(x^2)^2
- n^2 =(x^4) hence it has 4+1 factors
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