Prove by Induction!

Another doubt in induction, please help me out.

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

#NumberTheory #Induction

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

Step $1$ : 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 $n>0$ .

A Former Brilliant Member - 5 years, 8 months ago

Excellent, thank you genius!

Swapnil Das - 5 years, 8 months ago

Your Welcome :)

A Former Brilliant Member - 5 years, 8 months ago

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.

A Former Brilliant Member - 5 years, 8 months ago

That's an obvious hint lol xD

Nihar Mahajan - 5 years, 8 months ago

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

Swapnil Das - 5 years, 8 months ago

Only tool (with me ):MATHEMATICAL INDUCTION

Aakash Khandelwal - 5 years, 8 months ago

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 - 5 years, 8 months ago

Lol, I assumed powers of $5$ to be even by mistake, and thus couldn't get the result!

Swapnil Das - 5 years, 8 months ago

This is true only for n equals 1. It is not true for numbers bigger than 1.

mohamed sultan - 5 years, 7 months ago

@ Mohamed Sultan - You are wrong. Go back and try it for n = 2, 3, and 4, for instance.

Linda Slovik - 2 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$