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Math Help - " Help me in solving problem"

  1. #1
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    " Help me in solving problem"

    I did not understand the given problem statement.
    Could you please explain it clearly?
    And give me step by step answer

    The given problem is

    " Find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively."

    Thanking you , advance.

    Please explain in details.
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by SEK790 View Post
    I did not understand the given problem statement.
    Could you please explain it clearly?
    And give me step by step answer

    The given problem is

    " Find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively."

    Thanking you , advance.

    Please explain in details.
    Are you sure this problem should be in Pre-Algebra and Algebra? This is a problem for the Chinese Remainder Theorem.

    x\equiv 6\ (\text{mod}\ 1657)

    x\equiv 5\ (\text{mod}\ 2037)

    1657 and 2035 are pairwise coprime. Solving the CRT gives

    x \equiv 1305722\ (\text{mod}\ 3375309)

    So the answer is 1305722.

    EDIT: I misread "greatest number" as "smallest positive integer." There are infinitely many numbers satisfying the given constraints, and there is no greatest number.

    EDIT 2: Ah, it seems I made another misinterpretation. I thought you meant "dividing by" instead of "dividing." Okay so we have

    1657\equiv 6\ (\text{mod}\ x)

    2037\equiv 5\ (\text{mod}\ x)

    This leads to

    1651\equiv 0\ (\text{mod}\ x)

    2032\equiv 0\ (\text{mod}\ x)

    where

    1651 = 13\cdot 127

    2032 = 2^4\cdot 127

    So the answer is 127.
    Last edited by undefined; May 31st 2010 at 11:47 PM.
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  3. #3
    MHF Contributor undefined's Avatar
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    Since you haven't responded, and since the problem is easier than the problem I thought it was at first, I'll write a simpler explanation here.

    We are after a number, call it x.

    1657 divided by x gives remainder 6. That means that 1657 - 6 = 1651 is a multiple of x.

    2037 divided by x gives remainder 5. That means that 2037 - 5 = 2032 is a multiple of x.

    We take the greatest common divisor of 1651 and 2032, and to help with that I wrote the prime factorisations in my post above.

    If the greatest common divisor had been less than or equal to 6, this problem would have no solution.
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