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rational
 one year ago
Find the \(\gcd\) of the set of numbers
\[\{16^n+10n1n=1,2,3,\ldots\}\]
rational
 one year ago
Find the \(\gcd\) of the set of numbers \[\{16^n+10n1n=1,2,3,\ldots\}\]

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rational
 one year ago
Best ResponseYou've already chosen the best response.1I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Suppose 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.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0hmm do u have some other way ?

rational
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{align}\color{blue}{16^n}+10n1 &= \color{blue}{(1+5*3)^n}+10n1 \\~\\ &= \color{blue}{1+5*3n + 5^2k} + 10n1 \\~\\&= 25M\end{align}\]

rational
 one year ago
Best ResponseYou've already chosen the best response.1recall that \((1+x)^n = 1+nx+x^2(stuff)\) from binomial thm
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