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Math Help - Remainder Problem

  1. #1
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    Remainder Problem

    Hi All,
    Q If x and y are postive integers and x/y has a remainder of 5, what is the smallest possible value of xy?

    1. y > 5 as the dividor has to be larger than the remainder. so let's take 6.
    2. The answer goes on to explain 5 as the smallest value of x that can be divided by 6 and leave a remainder of 5. (But 5/6 does not leave a remainder of 5). 11, 17, 23 would though.

    Can anyone explain x being 5 to me? Much appreciated.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Yes, 5 would be.

    5/6 = 0

    remainder 5.

    It's just like you have 5 people and are requested to make teams of 6 people. How many teams of 6 people would you have and how many people will be left over?

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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    Yes, 5 would be.

    5/6 = 0

    remainder 5.

    It's just like you have 5 people and are requested to make teams of 6 people. How many teams of 6 people would you have and how many people will be left over?

    ah yes.. thanks. And the fact the question denotes 'remainder' would suggest you take this approach? Elsewhere 5/6 would equal: .833333.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    That's what remainder is all about.

    Take 11.

    11/6 = 1 and remainder 5, but it's also 1.8333

    It's the same thing here. It's not a different approach, you just are not used to the quotients being 0 with a remainder.
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