Another doubt in induction, please help me out.

Show that \(2\cdot7^{n} + 3\cdot5^{n} -5\) is divisible by 24.

Another doubt in induction, please help me out.

Show that \(2\cdot7^{n} + 3\cdot5^{n} -5\) is divisible by 24.

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TopNewestStep \(1\): when \(n=1\), we have \(2 \cdot 7^{n}+3 \cdot 5^{n}-5=2 \cdot 7+3 \cdot 5 -5=24\) which is divisible by \(24\).

Step \(2\): Assume this to be true for \(n=m\). Then \(2 \cdot 7^{m}+3 \cdot 5^{m}-5\) is divisible by \(24\).

Step \(3\): Now, \(2 \cdot 7^{m+1}+3 \cdot 5^{m+1}-5\)

\(=2 \cdot 7^{m} \times 7+ 3 \cdot 5^{m} \times 5-5\)

\(=14 \cdot 7^{m}+15 \cdot 5^{m}-5\)

\(=2 \cdot 7^{m}+3 \cdot 5^{m}-5+12 \cdot 7^{m}+12 \cdot 5^{m}\)

Since \(2 \cdot 7^{m}+3 \cdot 5^{m}-5\) is divisible by \(24\), we have only to prove that \(12 \cdot 7^{m}+12 \cdot 5^{m}\) is divisible by \(24\).

\(12 \cdot 7^{m}+12 \cdot 5^{m}=12(7^{m}+5^{m})\) which is divisible by \(24\) since \(7^{m}\) and \(5^{m}\) are both odd numbers and their sum is an even number.

Hence \(2 \cdot 7^{n}+3 \cdot 5^{n}-5\) is divisible by \(24\) for all integers \(n>0\). – Svatejas Shivakumar · 1 year, 1 month ago

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– Swapnil Das · 1 year, 1 month ago

Excellent, thank you genius!Log in to reply

– Svatejas Shivakumar · 1 year, 1 month ago

Your Welcome :)Log in to reply

Hint: Show that it is true for n=1. Then show that if the statement is true for any arbitrary positive integer m, then it is true for m+1. – Svatejas Shivakumar · 1 year, 1 month ago

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– Nihar Mahajan · 1 year, 1 month ago

That's an obvious hint lol xDLog in to reply

– Swapnil Das · 1 year, 1 month ago

Can you show the m+1 step? This is where I'm getting stuck at.Log in to reply

Only tool (with me ):MATHEMATICAL INDUCTION – Aakash Khandelwal · 1 year, 1 month ago

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I'm surprised that you don't know Induction. You should read the wiki and be familiar with it. This is a standard problem in divisibility by induction. – Calvin Lin Staff · 1 year, 1 month ago

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– Swapnil Das · 1 year, 1 month ago

Lol, I assumed powers of \(5\) to be even by mistake, and thus couldn't get the result!Log in to reply

This is true only for n equals 1. It is not true for numbers bigger than 1. – Mohamed Sultan · 1 year, 1 month ago

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