n3 = 450 y

n3 = 9 x 5 x 10 y

n3 = 3 x 3 x 5 x 5 x 2 y

Because n is in its 3rd power.

This implies that

Y has to have a 3, a 5, two 2’s and may be other

(lets call it x3) as its prime factors

Thus,

Y = 3 * 5 * 2 * 2 * x3Condn 1->y/ 3*4*5

Replacing Y with its factors

(3 * 5 * 2 * 2 * x3) / 3*4*5

Nothing left in the denominator -> Must be Integer

I is correct.

[b]Condn 2->[/b]

y/ 3*3*2*5

Replacing Y with its factors

(3 * 5 * 2 * 2 * x3) / 3*3*2*5

A 3 left in the denominator ->

May be Integer II is not correct.

Condn 3->y/ 3*2*5*5

Replacing Y with its factors

(3 * 5 * 2 * 2 * x3) / 3*2*5*5

A 5 left in the denominator ->

May be Integer III is not correct.

Ans is

B
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