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# m^2+n^2=5k+b ; b=0? I. m-n=5z+1 II. m+n=5t+3

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Director
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m^2+n^2=5k+b ; b=0? I. m-n=5z+1 II. m+n=5t+3 [#permalink]

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19 Nov 2008, 13:54
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$$m^2+n^2=5k+b$$; $$b=0?$$

I. $$m-n=5z+1$$
II. $$m+n=5t+3$$

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Intern
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19 Nov 2008, 18:26
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.
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Senior Manager
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20 Nov 2008, 01:56
C.
Masuhari , ur approach is right, but u forgot to square right side of the equation.
IF U had done so, u would have come to b=0.

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Director
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20 Nov 2008, 04:01
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.

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20 Nov 2008, 04:31
I will simply mark E as we do not know whether these variables are integer or fraction.

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20 Nov 2008, 10:13
scthakur wrote:
I will simply mark E as we do not know whether these variables are integer or fraction.

OA is not E though

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21 Nov 2008, 02:23
1
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

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21 Nov 2008, 05:36
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.
....

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22 Nov 2008, 19:33
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!

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22 Nov 2008, 22:53
ngotuan wrote:
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.
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Director
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30 Nov 2008, 12:23
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.

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Re: DS: divisibility   [#permalink] 30 Nov 2008, 12:23
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