The number of divisors of the form (4n+2...

The number of divisors of the form `(4n+2)` of the integer 240 is

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We have,
`4m+2and240=2^(4)xx3xx5`
Therefore, in any divisor of the form 4m+2, number 2 must occur exactly once. Hence, the number of divisors of the form 4m+2 is
`1xx(1+1)(1+1)=4`

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OBJECTIVE RD SHARMA ENGLISH-PERMUTATIONS AND COMBINATIONS-Chapter Test

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