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Re: DS: divisibility [#permalink]
20 Nov 2008, 03:01

masuhari wrote:

this one puzzles me...but I'd pick C

Stmt 1 or 2 by themselves not sufficient..coz you don't know the value of 2mn, when you square both of the LHSs.

Squaring both and adding them results in m^2+n^2-2mn = 5z + 1 m^2+n^2+2mn = 5t + 3 OR 2m^2+2n^2 = 5(z+t) + 4

simplifying, we get m^2+n^2 = 5(z+t)/2 + 2

Looks like we can conclude if b = 0 or not from that. Hence C.

masuhari, if you correctly square RHS, eventually we will have (25/2)(z^2+t^2) as one of the expressions. How can one know whether this expression is divisible by 5? To be able to say "Yes" or "No" we would need to know whether z^2+t^2 is even.

Re: DS: divisibility [#permalink]
21 Nov 2008, 01:23

1

This post received KUDOS

From: m-n=5z+1 m+n=5t+3 we have : 2m=5(z+t)+4 then (z+t) is divisible by 2 2n=5(t-z)+3 then (t-z) is divisible by 2 and then z,t is divisible by 2 or z=2p;t=2q Otherwise, from: m-n=5z+1 m+n=5t=3 we have 2(m^2+n^2)=25(z^2+t^2)+10z+30t+10 or m^2+n^2=25(z^2+t^2)/2+5z+15t+5 we can see that 25(z^2+t^2)/2=25(4p^2+4q^2)/2 is an integer so (m^2+n^2)=5.land we can conclude that b=0

Re: DS: divisibility [#permalink]
21 Nov 2008, 04:36

ngotuan wrote:

From: m-n=5z+1 m+n=5t+3 we have : 2m=5(z+t)+4 then (z+t) is divisible by 2 2n=5(t-z)+3 then (t-z) is divisible by 2 and then z,t is divisible by 2 or z=2p;t=2q Otherwise, from: m-n=5z+1 m+n=5t=3 we have 2(m^2+n^2)=25(z^2+t^2)+10z+30t+10 or m^2+n^2=25(z^2+t^2)/2+5z+15t+5 we can see that 25(z^2+t^2)/2=25(4p^2+4q^2)/2 is an integer so (m^2+n^2)=5.land we can conclude that b=0

I have one doubt ngotuan, t+z is divisible by 2 t-z is divisible by 2 but it does not mean..that t is divisible by 2. for example, take t=5, z=9 t+z=14 divisible by 2 and t-z=4 divisible by 2 but t & z are not divisible by 2. ....

Re: DS: divisibility [#permalink]
22 Nov 2008, 18:33

oops,sorry.I made a mistake. We can resolve this : m^2+n^2=25(z^2+t^2)/2+5z+15t+5 we need to sure that (z^2+t^2)is divisible by 2. as I mentioned, (t-z) and (t+z) are divisible by 2 , so (t-z)(t+z) is divisible by 2. We have : t^2-z^2 is divisible by 2. Otherwise (z^2+t^2)=(t^2-z^2)+2z^2 is divisible by 2---> from this we have (m^2+n^2)=5.l Sorry again!

Re: DS: divisibility [#permalink]
22 Nov 2008, 21:53

ngotuan wrote:

oops,sorry.I made a mistake. We can resolve this : m^2+n^2=25(z^2+t^2)/2+5z+15t+5 we need to sure that (z^2+t^2)is divisible by 2. as I mentioned, (t-z) and (t+z) are divisible by 2 , so (t-z)(t+z) is divisible by 2. We have : t^2-z^2 is divisible by 2. Otherwise (z^2+t^2)=(t^2-z^2)+2z^2 is divisible by 2---> from this we have (m^2+n^2)=5.l Sorry again!

Why (t-z) or (t+z) has to be divisible by 2? & How are they divisible by 2?

ok, 2m = 5(t+z) + 4 but do not know that m or n is an integer.
_________________

Re: DS: divisibility [#permalink]
30 Nov 2008, 11:23

Guys, OA C. Thanks for a try.

I have just realised that for 2(m^2+n^2)=25(z^2+t^2)+10z+30t+10 to make sense z^2+t^2 needs to be even, otherwise RHS will be odd which would be incorrect given LHS is even.

gmatclubot

Re: DS: divisibility
[#permalink]
30 Nov 2008, 11:23