If $37^{n + 2} + 16^{n + 1} + 30^{n}$ is divided by $7$ is

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Solution

The correct option is A $0$
$37^{n + 2} = (7 \times 5 + 2)^{2 n + 2}$
$= 7 A + 2^{n + 2}$ where K is positive integer.
Similarly
$16^{n + 1} = (7 \times 2 + 2)^{n + 1} = 7 B + 2^{n + 1}$
$30^{n} = (7 \times 4 + 2)^{n} = 7 C + 2^{n}$
Hence
$7 A + 2^{n + 2} + 7 B + 2^{n + 1} + 7 C + 2^{n}$
$= 7 (A + B + C) + 2^{n} (4 + 2 + 1)$
$= 7 (A + B + C + 2^{n})$
Which is divisible by $7$ .

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If 37n+2+16n+1+30n is divided by 7 is

The correct option is A 037n+2=(7×5+2)2n+2=7A+2n+2 where K is positive integer.Similarly 16n+1=(7×2+2)n+1=7B+2n+130n=(7×4+2)n=7C+2nHence7A+2n+2+7B+2n+1+7C+2n=7(A+B+C)+2n(4+2+1)=7(A+B+C+2n)Which is divisible by 7.

If $37^{n + 2} + 16^{n + 1} + 30^{n}$ is divided by $7$ is