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another one

This topic has 3 member replies
iamtrying Newbie | Next Rank: 10 Posts Default Avatar
Joined
21 Aug 2006
Posted:
8 messages

another one

Post Tue Oct 17, 2006 9:29 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    . If M and N are positive integers that have remainders of 1 and 3, respectively, when divided by 6, which of the following could NOT be a possible value of M+N?
    (A) 86
    (B) 52
    (c)34
    d)28
    e)10

    i am thinking answer is (E). Correct?

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    countingdolls Junior | Next Rank: 30 Posts Default Avatar
    Joined
    02 Oct 2006
    Posted:
    13 messages
    Post Tue Oct 17, 2006 10:12 am
    No.
    (A)

    take M as (6n + 1), N as (6n+3).
    so, M+N= 2(6n) + 4.

    now all you have to do is choose that option which isnt greater than a multiple of 6 by 4.So, 86.

    anandsebastin Junior | Next Rank: 30 Posts Default Avatar
    Joined
    13 Oct 2006
    Posted:
    10 messages
    Post Wed Oct 18, 2006 6:35 pm
    iamtrying wrote:
    . If M and N are positive integers that have remainders of 1 and 3, respectively, when divided by 6, which of the following could NOT be a possible value of M+N?
    (A) 86
    (B) 52
    (c)34
    d)28
    e)10

    i am thinking answer is (E). Correct?
    I'm afraid not. Answer is A.

    M+N = (6x+1)+(6y+3)
    =6(x+y)+4

    So, all possible values of M+N should satisfy, (M+N) - 4/6 = integer.

    Applying this test, you'll see that B,C,D and E satisfy this equation.

    As for your conclusion, when you divide 3 by 6, remainder is 3, right? Wink

    veer_jy Newbie | Next Rank: 10 Posts Default Avatar
    Joined
    18 Oct 2006
    Posted:
    2 messages
    Post Wed Oct 18, 2006 7:38 pm
    (A) sounds correct!

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