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  1. #1
    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 $\displaystyle 8^n - 7n - 1 \equiv 0\ (\text{mod }49)$

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

    Use mathematical induction. Base case, n = 1.

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

    Induction step.

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

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

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


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

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

    . . . . . . . . . $\displaystyle =\;\bigg[7^n + {n\choose n-1}7^{n-1} + {n\choose n-2}7^{n-2} + \hdots + {n\choose2}7^2 + {n\choose1}7 + 1
    \bigg] - 7n - 1$

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

    . . . . . . . . . $\displaystyle =\;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: .$\displaystyle x$ is a multiple of 49.

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