Question

Is 2352 a perfect square? if not find smaller multiple 2352 which is a perfect square.

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Answer 1

Answer

We have 2352=2×2×2×2×3×7×7

As the prime factor 3 has no pair, 2352 is not a perfect square, So we multiply 2352 by 3 to get, 2352 ×3

$= 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 7$

Now, each prime factor is a pair, therefore, 2352 ×3 =7056 is a perfect square.

Thus, the required smallest multiple of 2352 is 7056 which is a perfect square and

$\sqrt[]{7056} = 2 \times 2 \times 3 \times 7 = 84$

Answer 2

Method used - Prime Factorisation :

Prime factorisation:

Definition: A method of finding the the prime factors of the given number.

Uses: Finding of a number is a perfect square, cube, its prime factors etc.

Prime factorisation of 2352:

$2352 = {2}^{4} \times {3}^{1} \times {7}^{2}$

2352 is not a perfect square as a perfect square should have all of its factors in powers of 2 or its multiples.

But in this case 3 has a power of 1. So, we should multiply the number by such a number such that 3 has an even power. The smallest number by which it should be multiplied for it to become a perfect square is 3 as when we multiply it by 3, all its factors have an even power.

$2352 \times 3 = {2}^{4} \times {3}^{2} \times {7}^{2} \\ 7056 ={2}^{4} \times {3}^{2} \times {7}^{2} \\ \sqrt{7056} = \sqrt{ {2}^{4} \times {3}^{2} \times {7}^{2} } \\ \sqrt{7056} = {2}^{2} \times 3 \times 7 \\ \sqrt{7056} = 4 \times 21 \\ \sqrt{7056} = 84$

7056 is the square of 84.

Therefore, 7056 is the smallest multiple of 2352 which is a perfect square.