Question

Find the \(\gcd\) of the set of numbers
\[\{16^n+10n-1|n=1,2,3,\ldots\}\]

Answer 1

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

Answer 2

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.

Answer 3

Thanks!

Answer 4

hmm do u have some other way ?

Answer 5

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

Answer 6

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

Answer 7

got that :)