Question

Number of divisors of the form
4n + 2 (n ≥ 0) of the integer 240 is

(a) 4

(b) 8

(c) 10

(d) 3

❌❌ NO SPAMMING❌❌​

Answer 1

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Option (a)

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$<b>Firstly,$

factors of 240 are-

$240 = {2}^{4} \times 3 \times 5$

Given that divisor is of the form-

$= 4n + 2 \\ = 2(2n + 1)$

$2(2n + 1) \: shows \: that \: it \: is \: \\ divisible \: by \: 2 \: but \\ not \: by \: 4$

The only divisors that we need to consider are those with a maximum power of 2^1.

The divisors of this form are,

$=(2) \: \: \: (2 \times 3) \: \: \:( 2 \times 5) \\ \: \: \:and \: ( 2 \times 3 \times 5)$

$<b><u>Hence, there are 4 such numbers.$

Answer 2

240= 2^4 × 3 × 5

4n +2 = 2( 2n +1)

2n +1 is odd

So1, 3, 5 , 3× 5 these are odd

So 4 divisors