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    Member grgrsanjay's Avatar
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    Post sets

    if x={8^n - 7n - 1} and y={49(n-1)} prove that x is a subset of y
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    Quote Originally Posted by grgrsanjay View Post
    if x={8^n - 7n - 1} and y={49(n-1)} prove that x is a subset of y
    So presumably n is a positive integer.

    You need to show that 8^n - 7n - 1 \equiv 0\ (\text{mod }49)

    (The expression also must be non-negative but this is trivial.)

    Use mathematical induction. Base case, n = 1.

    8^1 - 7(1) - 1 = 0 \equiv 0\ (\text{mod }49)

    Induction step.

    8^{n+1} - 7(n+1) - 1 \equiv 8 \cdot 8^n - 7n - 7 - 1 \equiv (8)(7n+1) - 7n - 8 \equiv
     56n + 8 - 7n - 8 \equiv 49n \equiv  0\  (\text{mod }49)
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  3. #3
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    Hello, grgrsanjay!

    If x\:=\:8^n - 7n - 1\,\text{ and }\,y\:=\:49(n-1), prove that: . x \subset y.

    We want to show that x \:=\:8^n - 7n - 1 is always a multiple of 49.


    We have: . x\;=\;8^n - 7n - 1

    . . . . . . . . . =\;(7+1)^n - 7n - 1

    . . . . . . . . . =\;\bigg[7^n + {n\choose n-1}7^{n-1} + {n\choose n-2}7^{n-2} + \hdots + {n\choose2}7^2 + {n\choose1}7 + 1 <br />
\bigg] - 7n - 1

    . . . . . . . . . =\;7^n + {n\choose n-1}7^{n-1} + {n\choose n-2}7^{n-2} + \hdots + {n\choose2}7^2

    . . . . . . . . . =\;7^2\underbrace{\bigg[7^{n-2} + {n\choose n-1}7^{n-3} + {n\choose n-2}7^{n-4} + \hdots + {n\choose2}\bigg]}_{\text{This is an integer}}


    Therefore: . x is a multiple of 49.

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