JEE Main 2016 (19)

$37^{n+2} + 16^{n+1} + 30^{n}$

$= (35+2)^{n+2} + (14+2)^{n+1} + (28+2)^{n}$

In the binomial expansion of each of the above terms, all but the last terms are multiples of 7. So we only need to check divisibility of the last terms of each of the expansions. The last terms are:

$= {n+2 \choose n+2}2^{n+2} +{n+1 \choose n+1}2^{n+1} + {n\choose n} 2^n$

$= 2^{n+2} + 2^{n+1} + 2^n$

$= 2^n\times ( 2^2 +2^1 + 2^0)$

$= 2^n \times 7$

Hence, the sum of the these is also a multiple of 7. So the remainder when the given expression is divided by 7 is $\boxed{0}$ .

Nt say

$\large{ 37^{n+2}+16^{n+1}+30^{n} \equiv 2^{n+2}+2^{n+1}+2^n = 2^n(4+2+1)=2^n \times 7 \equiv 0 \pmod{7}}$

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Here $a \equiv b \pmod{c}$ means a leaves same reminder as b when divided by c.