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Question

The total number of divisors of the form 4n+2(n0) of integer 240 is

A
2
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B
4
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C
6
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D
8
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Solution

The correct option is B 4
240=24×3×5
4n+2=2(2n+1)=2×odd
the required number of divisors
= the number of selections of one 2 from four 2s, any number of 3s from one 3 and any number of 5s from one 5.
=1×(1+1)×(1+1)=1×2×2=4

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