Question

Using the prime factorisation method find which
of the following are perfect square?
(a) 51984 (b) 343 (c) 140250
(d) 6561 (e) 2592
51984​

Answer 1

Given:

(a) 51984 (b) 343 (c) 140250  (d) 6561 (e) 2592

To find:

Using the prime factorization method find which  of the following are perfect square?

Solution:

Using prime factorization method, we will solve the given problem as follows:

(a) 51984

Factors of 51984 = 2 × 2 × 2 × 2 × 3 × 3 × 19 × 19

∴ $\sqrt{51984} =$ $\sqrt{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 19 \times 19}$ = $2 \times 2 \times 3 \times 19 = 228$

Thus, 51984 is a perfect square.

(b) 343

Factors of 343 = 7 × 7 × 7

∴ $\sqrt{343} =$ $\sqrt{7 \times 7 \times 7}$ = $7 \sqrt{7}$

Thus, 343 is not a perfect square.

(c) 140250

Factors of 140250 = 2 × 3 × 5 × 5 × 5 × 11 × 17

∴ $\sqrt{140250} =$ $\sqrt{2 \times 3\times 5\times 5\times5\times 11\times17}$ = $5 \sqrt{5610}$ = $374.49$

Thus, 140250 is not a perfect square.

(d) 6561

Factors of 6561 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3

∴ $\sqrt{6561} =$ $\sqrt{3 \times 3\times 3\times 3\times3\times 3\times3\times 3}$ = $3\times3\times3\times 3$ = $81$

Thus, 6561 is a perfect square.

(e) 2592

Factors of 2592 = 2 × 2 × 2 × 2 × 2 × 3 × 3  × 3 × 3

∴ $\sqrt{2592} =$ $\sqrt{2 \times 2\times 2\times 2\times2\times 3\times3\times3\times3}$ = $36 \sqrt{2}$

Thus, 2592 is not a perfect square.

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Answer 2

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