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If $49^{n}+16 n+P$ is divisible by 64 for all $n \in N$, then the least negative integral value of $P$ is
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Verified Answer
The correct answer is:
$-1$
Put $n=1$ in $49^{n}+16 n+P$, then
$$
(49)^{1}+16(1)+P=49+16+P=65+P
$$
So, here if $P=-1$
Then, the given expression is divisible by 64 .
$$
(49)^{1}+16(1)+P=49+16+P=65+P
$$
So, here if $P=-1$
Then, the given expression is divisible by 64 .
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