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Q. Number of divisors of the form $(4n + 2), n\ge 0$ of the integer $240$ is

IIT JEEIIT JEE 1998Permutations and Combinations

Solution:

Since, $240 = 2^4 \cdot 3 \cdot 5$
$\therefore $Total number of divisors $= (4 + 1)(2)(2) =20$
Out of these $2, 6, 10$, and $30$ are of the form $4n + 2$.
Therefore, (a) is the answer.