Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 02 Jul 2015, 22:17

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Director
Joined: 29 Aug 2005
Posts: 879
Followers: 7

Kudos [?]: 213 [0], given: 7

m^2+n^2=5k+b ; b=0? I. m-n=5z+1 II. m+n=5t+3 [#permalink]  19 Nov 2008, 12:54
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
$$m^2+n^2=5k+b$$; $$b=0?$$

I. $$m-n=5z+1$$
II. $$m+n=5t+3$$
Intern
Joined: 14 Sep 2003
Posts: 45
Location: california
Followers: 1

Kudos [?]: 15 [0], given: 0

Re: DS: divisibility [#permalink]  19 Nov 2008, 17:26
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.
_________________

excellence is the gradual result of always striving to do better

Senior Manager
Joined: 28 Feb 2007
Posts: 306
Followers: 1

Kudos [?]: 32 [0], given: 0

Re: DS: divisibility [#permalink]  20 Nov 2008, 00:56
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.
Director
Joined: 29 Aug 2005
Posts: 879
Followers: 7

Kudos [?]: 213 [0], given: 7

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.
SVP
Joined: 17 Jun 2008
Posts: 1570
Followers: 12

Kudos [?]: 203 [0], given: 0

Re: DS: divisibility [#permalink]  20 Nov 2008, 03:31
I will simply mark E as we do not know whether these variables are integer or fraction.
Director
Joined: 29 Aug 2005
Posts: 879
Followers: 7

Kudos [?]: 213 [0], given: 7

Re: DS: divisibility [#permalink]  20 Nov 2008, 09:13
scthakur wrote:
I will simply mark E as we do not know whether these variables are integer or fraction.

OA is not E though
Intern
Joined: 26 Sep 2008
Posts: 19
Followers: 0

Kudos [?]: 5 [1] , given: 0

Re: DS: divisibility [#permalink]  21 Nov 2008, 01:23
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
Manager
Joined: 14 Nov 2008
Posts: 199
Schools: Stanford...Wait, I will come!!!
Followers: 3

Kudos [?]: 44 [0], given: 3

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.
....
Intern
Joined: 26 Sep 2008
Posts: 19
Followers: 0

Kudos [?]: 5 [0], given: 0

Re: DS: divisibility [#permalink]  22 Nov 2008, 18:33
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!
SVP
Joined: 29 Aug 2007
Posts: 2493
Followers: 59

Kudos [?]: 576 [0], given: 19

Re: DS: divisibility [#permalink]  22 Nov 2008, 21:53
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.
_________________
Director
Joined: 29 Aug 2005
Posts: 879
Followers: 7

Kudos [?]: 213 [0], given: 7

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.
Re: DS: divisibility   [#permalink] 30 Nov 2008, 11:23
Similar topics Replies Last post
Similar
Topics:
Is it true that exactly two of the three sentences I, II, 5 02 Nov 2011, 06:18
Is X + Y Negative? i) X is Positive. ii) Y is Negative. It's 6 15 May 2011, 05:42
If ax + b = 0, is x > 0 (i) a + b > 0 (ii) a - b < 5 24 Apr 2008, 15:05
Is the integer n odd? ( i ) n is divisible by 3. ( ii ) 2n 10 10 Jun 2007, 00:05
Is ab an even number ? I. a is divisible by 3. II. (b+1) is 4 23 Jun 2006, 15:58
Display posts from previous: Sort by