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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 \((m^2+n^2)=5.l\)and 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 \((m^2+n^2)=5.l\)and 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