Question

For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square number. Also find the square root of the square number so obtained.

1. 252
2. 2925
3. 396
4. 2645
5. 2800
6. 1620

Answer 1

(i) $252$ can be factorised as follows.

2	252
2	126
3	63
3	21
7	7
1

$252=\underline{2\times2}\times\underline{3\times3}\times7$
Here, prime factor $7$ does not have its pair.
If we divide this number by $7$, then the number will become a perfect square.
Therefore, $252$ has to be divided by $7$ to obtain a perfect square.
$\frac{252}{7}=36$ is a perfect square.
$36=\underline{2\times2}\times\underline{3\times3}$
∴ $\sqrt{36}=2\times3=6$

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(ii) $2925$ can be factorised as follows.

3	2975
3	925
5	325
5	65
13	13
1

$2925=\underline{3\times3}\times\underline{5\times5}\times13$
Here, prime factor $13$ does not have its pair.
If we divide this number by $13$, then the number will become a perfect square.
Therefore, $2925$ has to be divided by $13$ to obtain a perfect square.
$\frac{2925}{13} = 225$ is a perfect square.
$225=\underline{3\times3}\times\underline{5\times5}$
∴$\sqrt{225}=3\times5=15$

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(iii)$396$ can be factorised as follows

2	396
2	198
3	99
3	33
11	11
1

$396=\underline{2\times2}\times\underline{3\times3}\times11$
Here, prime factor $11$ does not have its pair.
If we divide this number by $11$, then the number will become a perfect square.
Therefore, $396$ has to be divided by $11$to obtain a perfect square.
$\frac{396 }{11} = 36$ is a perfect square.
$36=\underline{2\times2}\times\underline{3\times3}$
∴$\sqrt{36}=2\times3=6$

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(iv) $2645$ can be factorised as follows.

5	2645
23	529
23	23
1

$2645 = 5 \times \underline{23 \times 23 }$
Here, prime factor $5$ has no pair.
Therefore 2645 must be divided by $5$ to make it a perfect square.
$\sqrt{529} = 23 \times 23 = 23$