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. 
\(36=\underline{2\times2}\times\underline{3\times3}\)
∴ \(\sqrt{36}=2\times3=6\)

---

(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.
\(225=\underline{3\times3}\times\underline{5\times5}\)
∴\( \sqrt{225}=3\times5=15\)

---

(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.
\(36=\underline{2\times2}\times\underline{3\times3}\)
∴\(\sqrt{36}=2\times3=6\)

---

(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. 
\(\sqrt{529} = 23 \times 23 = 23\)