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If a and b are integers, is b even? (1) 3a + 4b is even. (2)

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If a and b are integers, is b even? (1) 3a + 4b is even. (2) [#permalink]  24 Dec 2005, 21:43
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50% (00:00) correct 50% (00:06) wrong based on 5 sessions
If a and b are integers, is b even?
(1) 3a + 4b is even.
(2) 3a + 5b is even.

I got the answer, but it took some time working out! Wanted to check any shorter methods!
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JAI HIND!

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Joined: 06 Jun 2004
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Re: DS: Even integer [#permalink]  24 Dec 2005, 22:51
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JAI HIND wrote:
If a and b are integers, is b even?
(1) 3a + 4b is even.
(2) 3a + 5b is even.

I got the answer, but it took some time working out! Wanted to check any shorter methods!

For a sum of two numbers to be even..two numbers can either

EVEN + EVEN
ODD + ODD

(1) 3a + 4b is even ==> 4b cannot be odd so it must be the even + even combination. B could be odd or even ==> Insufficient

(2) 3a + 5b is even ==> Could be EVEn + EVEN, ODD + ODD combo. Insufficient

(1) + (2) ==> From (1), we know that 3a must be even...so in (2) 5b must be even as well ==> b = even
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Joined: 20 Nov 2005
Posts: 2910
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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Re: DS: Even integer [#permalink]  25 Dec 2005, 00:14
TeHCM wrote:
JAI HIND wrote:
If a and b are integers, is b even?
(1) 3a + 4b is even.
(2) 3a + 5b is even.

I got the answer, but it took some time working out! Wanted to check any shorter methods!

For a sum of two numbers to be even..two numbers can either

EVEN + EVEN
ODD + ODD

(1) 3a + 4b is even ==> 4b cannot be odd so it must be the even + even combination. B could be odd or even ==> Insufficient

(2) 3a + 5b is even ==> Could be EVEn + EVEN, ODD + ODD combo. Insufficient

(1) + (2) ==> From (1), we know that 3a must be even...so in (2) 5b must be even as well ==> b = even

I got C. I did the same way as done by TeHCM
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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Joined: 14 Apr 2005
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Location: India, Chennai
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Re: DS: Even integer [#permalink]  25 Dec 2005, 22:50
JAI HIND wrote:
If a and b are integers, is b even?
(1) 3a + 4b is even.
(2) 3a + 5b is even.

I got the answer, but it took some time working out! Wanted to check any shorter methods!

From 1 we get 3a+4b = 0 mod 2
From 2 we get 3a+5b = 0 mod 2

Subtracting 2 from 1 we get b = 0 mod 2 hence C.
Re: DS: Even integer   [#permalink] 25 Dec 2005, 22:50
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