( 22 ! × P ) is a perfect square. Find the least value of P, given that P belongs to a set of natural numbers. Options: A . &nbsp176358 B . &nbsp347687 C . &nbsp909095 D . &nbsp686769 E . &nbsp0

Answer: Option A : A 22! can be written as 2 19 × 3 9 × 5 4 × 7 3 × 11 2 × 13 × 17 × 19 . To make it a perfect square, we need to multiply it by 2 × 3 × 7 × 13 × 17 × 19 . So, P = 176358.

(22!×P) is a perfect square. Find the least value of P, given that P belong

Question

(22! \times P)

is a perfect square. Find the least value of P, given that P belongs to a set of natural numbers.

Answer: Option A
:
A
22! can be written as

2^{19} \times 3^{9} \times 5^{4} \times 7^{3} \times 11^{2} \times 13 \times 17 \times 19

.
To make it a perfect square, we need to multiply it by

2 \times 3 \times 7 \times 13 \times 17 \times 19

.
So, P = 176358.

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