# Thread: Show 24 divides (2*7^n + 3*5^n - 5) for all positive integers n, but ...

1. ## Show 24 divides (2*7^n + 3*5^n - 5) for all positive integers n, but ...

without using mathematical induction and also explicitly not using the application of mod in your demonstration.

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$\displaystyle Show \ \ 24 \ \ divides \ \ (2*7^n \ + \ 3*5^n \ - \ 5) \ \$ for all positive integers n.

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2. ## Re: Show 24 divides (2*7^n + 3*5^n - 5) for all positive integers n, but ...

We can observe that:

$\displaystyle \left(2\cdot7^{1}+3\cdot5^{n}-5\right)=24$

$\displaystyle \left(2\cdot7^{n+1}+3\cdot5^{n+1}-5\right)-\left(2\cdot7^{n}+3\cdot5^{n}-5\right)=12\left(5^n+7^n\right)$

So, we need to show that $\displaystyle 5^n+7^n$ is even. The sum of two odd numbers is even.

3. ## Re: Show 24 divides (2*7^n + 3*5^n - 5) for all positive integers n, but ...

Originally Posted by MarkFL
We can observe that:

$\displaystyle \left(2\cdot7^{1}+3\cdot5^{n}-5\right)=24$
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You had a typo for the second exponent at that moment:

$\displaystyle \left(2\cdot7^{1}+3\cdot5^{1}-5\right)=24$

4. ## Re: Show 24 divides (2*7^n + 3*5^n - 5) for all positive integers n, but ...

Originally Posted by greg1313
You had a typo for the second exponent at that moment:

$\displaystyle \left(2\cdot7^{1}+3\cdot5^{1}-5\right)=24$
Yes, and I can no longer edit, but you got my meaning.