| View previous topic :: View next topic |
| Author |
Message |
nuku888 Just gettin' started!
Joined: 19 Apr 2007 Posts: 5
Thanks given: 0 Thanked 0 times in 0 posts
|
Posted: Thu Apr 19, 2007 5:29 pm Post subject: OG 11 math problem solving question |
|
|
Hello everyone,
I was wondering if anyone can help me with the explanation for question 241 in the OG11.
Question
If the integer n has exactly three positive divisors, including 1 and n how many positive divisors does n^2 have?
I read the explanation at the back but I don't understand if anyone can explain that would be great thanks! |
|
| Back to top |
|
 |
|
|
zaffar Rising GMAT Star
Joined: 29 Dec 2006 Posts: 49
Thanks given: 0 Thanked 0 times in 0 posts
|
Posted: Thu Apr 19, 2007 9:09 pm Post subject: |
|
|
i did it by picking numbers and there are only 2 numbers 4 (three divisors 1 2 4) & 9 (1 3 9). and their squares 16 (1,2,4,8,16) and 81 (1,3,9,27,81)
so 5 is the answer.
there may be someother method but i did it this way. i hope it helps. _________________ http://www.beatthegmat.com/viewtopic.php?t=1523 |
|
| Back to top |
|
 |
nuku888 Just gettin' started!
Joined: 19 Apr 2007 Posts: 5
Thanks given: 0 Thanked 0 times in 0 posts
|
Posted: Fri Apr 20, 2007 2:09 pm Post subject: |
|
|
| Thanks for your reply. Sorry just got one more questions they mention something about square root n is also an integer, so does that mean square root 4 = 2 that is why they said that it is also integer? |
|
| Back to top |
|
 |
sosaha Just gettin' started!
Joined: 21 Apr 2007 Posts: 1
Thanks given: 0 Thanked 0 times in 0 posts
Location: India
|
Posted: Sat Apr 21, 2007 10:33 pm Post subject: |
|
|
lets consider the number to be 'n'. since it has only one more divisor other than 1 and n, n must be a perfect sqare. so say n=x^2(x is a prime).
=> n^2 = x^4
Now divisors of x^4 are 1, x, x^2, x^3, x^4.
i.e 5 divisors. _________________ -Wish u all the best,
Soumyajit. |
|
| Back to top |
|
 |
|