1.

The number of divisors of the form `(4n+2)` of the integer 240 isA. 4B. 8C. 10D. 3

Answer» 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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