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Math Help - Divisibility by 7 question

  1. #1
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    Divisibility by 7 question

    Hi there! I'm having some big trouble with this:

    Let there be x, y \in \mathcal{Z}. Show that 10x + y is a multiple of 7 if, and only if x - 2y is a multiple of 7. Using the equivalence above, find a divisibility criteria for 7.

    So, I'd love to show you where I'm stuck, if it was not in the very beginning. Tips?
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  2. #2
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    Split into the two statements:

    (1) If x - 2y is a multiple of 7, then 10x + y is a multiple of 7 as well.

    Since x - 2y is a multiple of 7, then x - 2y = 7k_1 \ \Leftrightarrow \ x = 7k_1 + 2y for some k_1 \in \mathbb{Z}. Now let's plug this in: 10(7k_1 + 2y) + y = 70k_1 + 21y = 7 (10k_1 + 3y)

    Obviously, this is a multiple of 7 so we're done.

    ___________________________________

    (2) If 10x + y is a multiple of 7, then x - 2y is a multiple of 7 as well.

    Since 10x + y is a multiple of 7, then 10x + y = 7k_2 \ \Leftrightarrow \ y = 7k_2 - 10x for some k_2 \in \mathbb{Z}

    Plug this into x - 2y and the procedure is the same as above.
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    Hello, Rafael!

    I'll do part of it for you . . .


    Let there be x, y \in \mathcal{Z}.
    Show that 10x + y is a multiple of 7 if and only if x - 2y is a multiple of 7.

    We are given: . x - 2y \:=\:7a\,\text{ for some integer }a.

    Multiply both sides by 10: . 10x - 20y \:=\:70a

    \text{Add }21y\text{ to both sides: }\;10x + y \:=\:70a + 21y\:=\:7(10a + 3y)


    Therefore, 10x+y is a multiple of 7.

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  4. #4
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    Quote Originally Posted by Rafael Almeida View Post
    10x + y is a multiple of 7 if, and only if x - 2y is a multiple of 7. Using the equivalence above, find a divisibility criteria for 7.
    Suppose for example that you have a 5-digit number abcde. This is equal to 10*abcd + e. The equivalence shows that this is divisible by 7 if and only if abcd 2*e is divisible by 7. But this number is equal to 10*abc + (d2*e). So it is divisible by 7 if and only if abc 2(d2*e) is divisible by 7. But this number is equal to 10*ab + (c2*d+4*e). So it is divisible by 7 if and only if ab 2(c2*d+4*e) is divisible by 7.But this number is equal to 10*abc + (d2*e). So it is divisible by 7 if and only if abc 2(d2*e) is divisible by 7. But this number is equal to 10*a + (b2*c+4*d8*e). So it is divisible by 7 if and only if a 2(b2*c+4*d8*e) = a2*b+4*c8*d+16*e is divisible by 7.

    As an example, look at the number 42931. The criterion says that this should be divisible by 7 if and only if 42*2+4*98*3+16*1 is divisible by 7. But 42*2+4*98*3+16*1 = 28, which is divisible by 7. Therefore so is 42931.

    You can extend this criterion from 5-digit numbers to numbers with any number of digits in an obvious way.
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