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algebra!

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How many positive integers, a, are there such that 2027 divided by a leaves a remainder of 7?

  • algebra! - ,

    There is probably an easier way to go about doing this but I can't seem to think of it at the moment so you'll have to suffice with this until someone else can answer:

    I came up with this formula:
    a x b + 7 = 2027 where a is the the divisor and b is the quotient and 7 is the remainder. We know that if we multiple the divisor and quotient and add the remainder we should get the dividend.

    a x b = 2020

    Now we must find all of the factors of 2020.

    1 x 2020 = 2020 - therefore, a = 2020
    2 x 1010 = 2020 - therefore, a = 1010
    4 x 505 = 2020 - therefore, a = 505
    5 x 404 = 2020 - therefore, a=404
    10 x 202 = 2020 - therefore, a = 10 or 202
    20 x 101 = 2020, therefore, a = 20 or 101

    So the # of + integers for a such that 2027 divided by a leaves a remainder of 7 would be 8.

    I'm not too sure if I did that correctly. Hopefully someone can verify this.

  • algebra! - ,

    IT is Absolutely right . The answer is 8

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