# Prove by Induction!

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

Note by Swapnil Das
3 years, 6 months ago

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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 $$n>0$$.

- 3 years, 6 months ago

Excellent, thank you genius!

- 3 years, 6 months ago

- 3 years, 6 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.

- 3 years, 6 months ago

That's an obvious hint lol xD

- 3 years, 6 months ago

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

- 3 years, 6 months ago

Only tool (with me ):MATHEMATICAL INDUCTION

- 3 years, 6 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.

Staff - 3 years, 6 months ago

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

- 3 years, 6 months ago

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

- 3 years, 5 months ago

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

- 8 months, 3 weeks ago