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Number properties problem

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by kvcpk » Sun Jul 18, 2010 12:19 pm
daflower wrote:If n is a multiple of 5 where n=(p^2)(q) and p& q are both prime numbers, which is a multiple of 25?

a) p^2
b) q^2
c) pq
d) (p^2)(q^2)
e) (p^3)(q)

Can someone help explain how to solve?
n is a multiple o5. Let n=5k
n=(p^2)(q)
5k=(p^2)(q)

Since P and Q are both primenumbers, either p or q has to be multiple of 5.
If p is a multiple of 5, then p^2 is a multiple of 25.
If q is a multiple of 5, then q^2 is a multiple of 25.

Hence p^2q^2 shud be multiple of 25.

You can also do the same ny picking numbers.
let n=5, p=1, q=5 -> ACE out.
let n=25, p=5, q=1 -> B out

pick D
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by aloneontheedge » Sun Jul 18, 2010 12:31 pm
kvcpk wrote:
daflower wrote:If n is a multiple of 5 where n=(p^2)(q) and p& q are both prime numbers, which is a multiple of 25?

a) p^2
b) q^2
c) pq
d) (p^2)(q^2)
e) (p^3)(q)

Can someone help explain how to solve?
n is a multiple o5. Let n=5k
n=(p^2)(q)
5k=(p^2)(q)

Since P and Q are both primenumbers, either p or q has to be multiple of 5.
If p is a multiple of 5, then p^2 is a multiple of 25.
If q is a multiple of 5, then q^2 is a multiple of 25.

Hence p^2q^2 shud be multiple of 25.

You can also do the same ny picking numbers.
let n=5, p=1, q=5 -> ACE out.
let n=25, p=5, q=1 -> B out

pick D
Hey KVcpk,
p cannot be equal 1 as it is not a prime number
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by kvcpk » Sun Jul 18, 2010 12:35 pm
aloneontheedge wrote: Hey KVcpk,
p cannot be equal 1 as it is not a prime number
oops Sorry.. Thanks for the correction.. Its midnite here :)

I still think answer is correct. Need to put 2 instead of 1 in my examples.
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by aloneontheedge » Sun Jul 18, 2010 12:42 pm
kvcpk wrote:
aloneontheedge wrote: Hey KVcpk,
p cannot be equal 1 as it is not a prime number
oops Sorry.. Thanks for the correction.. Its midnite here :)

I still think answer is correct. Need to put 2 instead of 1 in my examples.
I agree...but i feel there is a better approach. Its 2 am here.
My mind is hardly working :-(
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by debmalya_dutta » Sun Jul 18, 2010 1:37 pm
kvcpk wrote:
daflower wrote:If n is a multiple of 5 where n=(p^2)(q) and p& q are both prime numbers, which is a multiple of 25?

a) p^2
b) q^2
c) pq
d) (p^2)(q^2)
e) (p^3)(q)

Can someone help explain how to solve?
n is a multiple o5. Let n=5k
n=(p^2)(q)
5k=(p^2)(q)

Since P and Q are both primenumbers, either p or q has to be multiple of 5.
If p is a multiple of 5, then p^2 is a multiple of 25.
If q is a multiple of 5, then q^2 is a multiple of 25.

Hence p^2q^2 shud be multiple of 25.

You can also do the same ny picking numbers.
let n=5, p=1, q=5 -> ACE out.
let n=25, p=5, q=1 -> B out

pick D
kvcpk -
Based on your solution , why isnt A the possible answer?
The question says (p^2)(q) is a mutliple of 5 and p and q are prime numbers ...
what if p = 5 and q =2?
p^2*q (= 25 * 2) is a multiple of 5.. p^2 is a multiple of 25.....
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by kvcpk » Sun Jul 18, 2010 8:55 pm
debmalya_dutta wrote: kvcpk -
Based on your solution , why isnt A the possible answer?
The question says (p^2)(q) is a mutliple of 5 and p and q are prime numbers ...
what if p = 5 and q =2?
p^2*q (= 25 * 2) is a multiple of 5.. p^2 is a multiple of 25.....
p^2 cannot be the answer because, p^2 is not always a multiple of 25.

Suppose n=45, then n= 3^2*5
Here p=3 and is a multiple of 9. not 25.

Goal here is to find a expression that is always a multiple of 25.

Hope this helps!!
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by outreach » Tue Jul 20, 2010 5:46 am
assumption: one of the prime number p or q will have value of 5
a. if p is not 5, then n cannot be multiple of 5
b. if q is not 5, then n cannot be multiple of 5
c. one possibility where this combination will pass his when both p and q are 5.
d. since one of the number has to be 5, when we square it, we will get n multiple of 25
e. if p is not 5, then n cannot be multiple of 5
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