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# 5.Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained. (i) 525 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412

(i)
$$\qquad \begin{array}{c|lcr} &22\\ \hline 2 &\;\; \;\bar{5}\bar{25}\\ &-4\\ \hline 42 &\;\;\; 125\\ &\;\;\; 84\\ \hline &\;\;\;41\\ \hline \end{array}$$

So we got remainder as 41.It represents that the square of 22 is less than 525.So, the next number is 23 and $$23^2 = 529$$.

Hence, the number to be added = 529 – 525 = 4

∴Required perfect square = 529

Thus we have, $$\sqrt{529}=23$$

(ii)
$$\qquad \begin{array}{c|lcr} &41\\ \hline 4 &\;\; \;\bar{17}\bar{50}\\ &-16\\ \hline 81 &\;\;\; 150\\ &\;\;\; 81\\ \hline &\;\;\;69\\ \hline \end{array}$$

So we got remainder as 69.It represents that the square of 41 is less than 1750.So, the next number is 42 and $$42^2 = 1764$$.

Hence, the number to be added = 1764– 1750 =14

∴Required perfect square = 1764

Thus we have, $$\sqrt{1764}=42$$

(iii)
$$\qquad \begin{array}{c|lcr} &15\\ \hline 1 &\;\; \;\bar{2}\bar{52}\\ &-16\\ \hline 25 &\;\;\; 152\\ &\;\;\; 125\\ \hline &\;\;\;27\\ \hline \end{array}$$

So we got remainder as 27.It represents that the square of 15 is less than 252.So, the next number is 16 and $$16^2 = 256$$.

Hence, the number to be added = 256 – 252 = 4

∴Required perfect square = 256

Thus we have, $$\sqrt{256}=16$$

(iv)
$$\qquad \begin{array}{c|lcr} &42\\ \hline 4 &\;\; \;\bar{18}\bar{25}\\ &-16\\ \hline 82 &\;\;\; 225\\ &\;\;\; 164\\ \hline &\;\;\;61\\ \hline \end{array}$$

So we got remainder as 61.It represents that the square of 42 is less than 1825.So, the next number is 43 and $$43^2 = 1849$$.

Hence, the number to be added = 1849 – 1825 = 24

∴New number = 1849

Thus we have,$$\sqrt{1849}=43$$

(v)
$$\qquad \begin{array}{c|lcr} &80\\ \hline 8 &\;\; \;\bar{64}\bar{12}\\ &-64\\ \hline 160 &\;\;\; 12\\ &\;\;\; 0\\ \hline &\;\;\;12\\ \hline \end{array}$$

So we got remainder as 12.It represents that the square of 80 is less than 6412.So, the next number is 81 and $$81^2 = 6561$$.

Hence, the number to be added = 6561 – 6412 = 149

∴New number = 6561

Thus, we have $$\sqrt{6561}=81$$