First slide

Combinations

Question

The number of divisors of 240 which are of the form $4 m + 2$ , is

Moderate

A

4
B

8
C

10
D

3

Solution

We have,
$4 m + 2 and 240 = 2^{4} \times 3 \times 5$
Therefore, in any divisor of the form 4m + 2, number 2 must occur exactly once. Hence, the number of divisors of the form $4 m + 2$ is
1X (1 + 1) (1 + 1) = 4

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