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Math Help - Division and remainder

  1. #1
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    Question Division and remainder

    What is the smallest number which when divided by 8,12 or 15 leaves remainder of 11 ?
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  2. #2
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    Re: Division and remainder

    Quote Originally Posted by haftakhan View Post
    What is the smallest number which when divided by 8,12 or 15 leaves remainder of 11 ?
    a number divided by 8 cannot have a remainder of 11
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    Re: Division and remainder

    But the question states this. Maybe the question is wrong
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  4. #4
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    Re: Division and remainder

    Hello, haftakhan!

    What is the smallest number which when divided by 8, 12 or 15
    leaves a remainder of 11?

    skeeter is absolutely right.
    I'll assume that the problem is poorly worded.

    The expected answer is probably:
    . . \text{LCM}(8,12,15) + 11 \:=\:120 + 11 \:=\:\boxed{131}


    Check

    . . 131 \div 15 \:=\:8,\;r\,11

    . . 131 \div 12 \:=\:10,\;r\,11

    And I suppose we can write:
    . . 131 \div 8 \:=\:15,\;r\,11

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  5. #5
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    Re: Division and remainder

    I assume you simply mean that x is some integer times 8, plus 11 (which would actually give a remainder of 11- 8= 3).

    Okay, let x be that number. Then we require that x- 8i= 11 for some integer, i, that x- 12j= 11 for some integer, j, and that x- 15k= 11 for some integer k. That last equation gives x= 11+ 15k.

    Putting x= 11+ 15k into the second equation, 11+ 15k- 12j= 11 so that 12j= 15k or j= \frac{5}{4}k. Since j must be an integer, k must be a multiple of 4.

    Putting x= 11+ 15k into the first equation, 11+ 15k- 8i= 11 so that 8i= 15k so i= \frac{15}{8}k. Since i must be an integer, k must be a multiple of 8. Since that is also a multiple of 4, the smallest value we can have for k is 8.

    That gives x= 11+ 15k= 11+ 15(8)= 11+ 120= 131.

    Note that, with k= 8, j= \frac{5}{4}(8)= 10. That gives x= 11+ 12(10)= 11+ 120= 131 again.

    Also, with k= 8, i= \frac{15}{8}(8)= 15. That gives x= 11+ 8(15)= 11+ 120= 131.

    As a final check, 131 divided by 8 is 16 with a remainder of 3 which could be interpreted as "15 with a remainder of 8+ 3= 11".
    131 divided by 12 is 10 with a remainder of 11.
    131 divided by 15 is 8 with a remainder of 11.

    Too slow again!
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