question_answer 1)

                For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained. (i) 252                                    (ii) 2925                (iii) 396                                  (iv) 2645 (v) 2800                                (vi) 1620

Answer:

                (i) 252 The prime factorisation of 252 is \[252=2\times 2\times 3\times 3\times 7\]. We see that the prime factor 7 has no pair. So, if we divide 252 by 7, then we get  

2	252
2	126
3	63
3	21
	7

\[252\div 7=\underline{2\times 2}\times \underline{3\times 3}\]. Now each prime factor has a pair. Therefore, 252 - 7 = 36 is a perfect square. Thus, the required smallest number is 7. Hence, \[\sqrt{36}=2\times 3=6\]. (ii) 2925 The prime factorisation of 2925 is\[2925=3\times 3\times 5\times 5\times 13\]. We see that the prime factor 13 has no pair. So, if we divide 2925 by 13, then we get

3	2925
3	975
5	325
5	65
	13

  \[2925\div 13=\underline{3\times 3}=\underline{5\times 5}\] Now each prime factor has a pair. Therefore, \[2925\div 13=225\] is a perfect square. Thus, the required smallest number is 13. Hence, \[\sqrt{225}\,=3\times 5=15\]. (iii) 396 The prime factorisation of 396 is\[396=2\times 2\times 3\times 3\times 11\] . We see that the prime factor 11 has no pair. So if we divide 396 by 11, then we get  

2	396
2	198
3	99
3	33
	11

\[396\div 11=\underline{2\times 2}\times \underline{3\times 3}\] Now each prime factor has a pair. Therefore, \[396\div 11=36\] is a perfect square. Thus the required smallest number is 11. Hence, \[\sqrt{36}\,=2\times 3=6\]. (iv) 2645 The prime factorisation of 2645 is \[2645=5\times 23\times 23\]. We see that the prime factor 5 has no pair. So, if we divide 2645 by 5, then we get

5	2645
23	529
	23

                \[2645\div 5=\underline{23\times 23}\] Now the only prime factor 23 has a pair. Therefore, \[2645\div 5=529\] is a perfect square. Thus, the required smallest number is 5. Hence, \[\sqrt{529}=23\]. (v) 2800 The prime factorisation of 2800 is \[2800=2\times 2\times 2\times 2\times 5\times 5\times 7\]. We see that the prime factor 7 has no pair. So, if we divide 2800 by 7, then we get

2	2800
2	1400
2	700
2	350
5	175
5	35
	7

  \[2800\div 7=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}\] Now each factor has a pair. Therefore, \[2800\div 7=400\] is a perfect square. Thus, the required smallest number is 7. Hence, \[\sqrt{400}=2\times 2\times 5=20\]. (vi) 1620 The prime factorisation of 1620 is \[1620=2\times 2\times 3\times 3\times 3\times 3\times 5\]. We see that the prime factor 5 has no pair. So, if we divide 1620 by 5, then we get

2	1620
2	810
3	405
3	135
3	45
3	15
	5

  \[1620\div 5=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\] Now each factor has a pair. Therefore, \[1620\div 5=324\] is a perfect square. Thus, the required smallest number is 5. Hence, \[\sqrt{324}=2\times 3\times 3=18\].