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Math Help - Please help

  1. #1
    Junior Member
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    Please help

    The product of any whole numbers each of which leaves a remainder of 1 on dividing by 8, also leaves a remainder of 1 on dividing by 8. Why?

    First I picked the any two numbers that leave a remainder of 1 on dividing by 8;
    9 and 17

    9 x 17 = 153 also leaves a remainder of 1 on dividing by 8.

    Then I did; 9 = 8 + 1; also 17 = 16 + 1

    and 9 x 17 = (8 + 1) x (16 + 1) = (8 x 19) + 1


    I am stuck on how to explain how this is so, not sure if there is another way of presenting this.

    Thank you for the help.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by fair_lady0072002 View Post
    The product of any whole numbers each of which leaves a remainder of 1 on dividing by 8, also leaves a remainder of 1 on dividing by 8. Why?

    First I picked the any two numbers that leave a remainder of 1 on dividing by 8;
    9 and 17

    9 x 17 = 153 also leaves a remainder of 1 on dividing by 8.

    Then I did; 9 = 8 + 1; also 17 = 16 + 1

    and 9 x 17 = (8 + 1) x (16 + 1) = (8 x 19) + 1


    I am stuck on how to explain how this is so, not sure if there is another way of presenting this.

    Thank you for the help.
    If a number leaves a remainder of 1 when divided by 8, then that number is equal to some other number times 8 plus 1.

    Take two number that have a remainder of 1 when divided by 8. If we let these two numbers be x and y, where a, b, x, y are all integers, we can let:
    x = 8a + 1
    y = 8b + 1

    Then the product of these numbers is:
    xy = (8a + 1)(8b + 1) = 64ab + 8a + 8b + 1 = 8(8ab + a + b) + 1

    Where (8ab + a + b) is another integer, we have that whatever x and y equal, if x and y have a remainder of 1 when divided by 8, then the product of x and y will have a remainder of 1 when divided by 8.
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  3. #3
    Junior Member
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    I appreciate your help. Thnk you.
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