Can someone show a step by step process on how to solve this problem.
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2Use parentheses, they are important. – Doug M Apr 22 '20 at 18:38
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https://math.stackexchange.com/questions/188657/why-is-an-bn-divisible-by-a-b – lab bhattacharjee Apr 22 '20 at 18:38
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That’s clearly false. – Vishu Apr 22 '20 at 18:39
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1$6n+1-1=6n$ and not all multiples of $6$ are multiplies of $5$. Are you sure you have the question written correctly? – CyclotomicField Apr 22 '20 at 18:45
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You really want to proof that $5$ is a divisor of $6$? – Fakemistake Apr 22 '20 at 19:26
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Welcome to Mathematics Stack Exchange. Did you mean $5$ divides $6^{n+1}-1$? – J. W. Tanner Apr 22 '20 at 19:29
2 Answers
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If you meant $5$ divides $6^{n+1}-1$, here's an outline for a proof by induction.
Base case: $n=0$: $5$ divides $6^1-1=5$.
Inductive step: Assume $5$ divides $6^{n}-1$.
Then can you see how to show that $5$ divides $6^{n+1}-1=6\times6^n-1=5\times6^n+6^n-1$?
J. W. Tanner
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No matter what, the statement is false in the present form (independent of possibly forgotten brackets) and in particular not provable by induction.
$5\not\mid 6\cdot 1 +1-1=6$
$5\not\mid 6(1 +1)-1=11$
$5\not\mid 6\cdot 1 +(1-1)=6$
Jonas Linssen
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