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If $\alpha$ and $\beta$ are the least positive integers such that for all $n \in N, n^3+\alpha n$ is divisible by 3 and $n^3-\beta n$ is divisible by 6 , then $\alpha+\beta=$
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The correct answer is:
$3$
If for all $n \in N, n^3+\alpha n$ is divisible by 3 , then
for $n=1$ also it is divisible by 3 , so the least positive integral value of $\alpha=2$
And, since for all $n \in N, n^3-\beta n$ is divisible by 6 , then for $n=1$ also it is divisible by 6 , so the least positive integral value of $\beta=1$.
So, $\alpha+\beta=2+1=3$.
Hence, option (b) is correct.
for $n=1$ also it is divisible by 3 , so the least positive integral value of $\alpha=2$
And, since for all $n \in N, n^3-\beta n$ is divisible by 6 , then for $n=1$ also it is divisible by 6 , so the least positive integral value of $\beta=1$.
So, $\alpha+\beta=2+1=3$.
Hence, option (b) is correct.
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