The number of positive divisors of 252 is

Question

The number of positive divisors of 252 is (A) 5 (B) 9 (C) 10 (D) 18

Tardigrade Admin · Accepted Answer

We know that, if a = p1α1 . p2α 2 ... Then, the total number of positive divisors of a is T (a) = (α1 + 1) (α2 + 1 )...... Given, 252 = 22 × 32 × 71 Here, α1 = 2, α2 = 2, α3 = 1 ∴ T (a) = (2 + 1) (2 + 1) (1 + 1) = 3 . 3 . 2 = 18

Q. The number of positive divisors of $252$ is

5

9

10

18

We know that, if $a = p_{1}^{α_{1}} . p_{2}^{α 2} ...$
Then, the total number of positive divisors of a
is $T (a) = (α_{1} + 1) (α_{2} + 1) ......$
Given, $252 = 2^{2} \times 3^{2} \times 7^{1}$
Here, $α_{1} = 2, α_{2} = 2, α_{3} = 1$
$∴ T (a) = (2 + 1) (2 + 1) (1 + 1)$
$= 3.3.2$
$= 18$

Q. The number of positive divisors of 252 is

5

9

10

18

We know that, if a=p1α1​​.p2α2​... Then, the total number of positive divisors of a is T(a)=(α1​+1)(α2​+1)...... Given, 252=22×32×71 Here, α1​=2,α2​=2,α3​=1 ∴T(a)=(2+1)(2+1)(1+1) =3.3.2 =18

Q. The number of positive divisors of $252$ is

We know that, if $a = p_{1}^{α_{1}} . p_{2}^{α 2} ...$
Then, the total number of positive divisors of a
is $T (a) = (α_{1} + 1) (α_{2} + 1) ......$
Given, $252 = 2^{2} \times 3^{2} \times 7^{1}$
Here, $α_{1} = 2, α_{2} = 2, α_{3} = 1$
$∴ T (a) = (2 + 1) (2 + 1) (1 + 1)$
$= 3.3.2$
$= 18$