## rational one year ago Find the $$\gcd$$ of the set of numbers $\{16^n+10n-1|n=1,2,3,\ldots\}$

1. rational

I see that $$25$$ is common by evaluating the given expression for first few values of $$n$$. Need to prove that $$25$$ factors out in every element

2. anonymous

Suppose 25 divides $$16^n + 10n -1$$. $$16^{n+1} + 10(n+1) -1 - (16^n + 10n -1) = 16^n.15 + 10$$ Clearly, 5 divides $$16^n.15 + 10$$. Dividing by 5, we have $$16^n.3+ 2$$ 16^n leaves remainder 1 when divided by 5, so 5 divides $$16^n.3+ 2$$ Therefore 25 divides$$16^n.15 + 10$$ So, 25 also divides $$16^{n+1} + 10(n+1) -1$$ . So by induction 25 divides every number in the set and is the required gcd.

3. rational

Thanks!

4. ikram002p

hmm do u have some other way ?

5. rational

\begin{align}\color{blue}{16^n}+10n-1 &= \color{blue}{(1+5*3)^n}+10n-1 \\~\\ &= \color{blue}{1+5*3n + 5^2k} + 10n-1 \\~\\&= 25M\end{align}

6. rational

recall that $$(1+x)^n = 1+nx+x^2(stuff)$$ from binomial thm

7. ikram002p

got that :)