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# 4.Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained. (i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000

(i)
$$\qquad \begin{array}{c|lcr} &20\\ \hline 2 &\;\; \;\bar{4}\bar{02}\\ &-4\\ \hline 4 &\;\;\; 02\\ \hline \end{array}$$

So we have got remainder as 2.So we can say that 2 is the smallest required number to be subtracted from 402 to get a perfect square.
Therefore, we have new number as :402-2=400

So, $$\sqrt{400}=20$$

(ii)
$$\qquad \begin{array}{c|lcr} &44\\ \hline 4 &\;\; \;\bar{19}\bar{89}\\ &-16\\ \hline 84 &\;\;\; 389\\ &\;\;\; 336\\ \hline &\;\;\;53\\ \hline \end{array}$$

So we have got remainder as 53.So we can say that 53 is the smallest required number to be subtracted from 1989 to get a perfect square.
Therefore, we have new number as :1989-53=1936

So, $$\sqrt{1936}=44$$

(iii)
$$\qquad \begin{array}{c|lcr} &57\\ \hline 5 &\;\; \;\bar{32}\bar{50}\\ &-25\\ \hline 107 &\;\;\; 750\\ &\;\;\; 749\\ \hline &\;\;\;1\\ \hline \end{array}$$

So we have got remainder as 1.So we can say that 1 is the smallest required number to be subtracted from 3250 to get a perfect square.
Therefore, we have new number as :3250-1=3249

So, $$\sqrt{3249}=57$$

(iv)
$$\qquad \begin{array}{c|lcr} &28\\ \hline 2 &\;\; \;\bar{8}\bar{25}\\ &-4\\ \hline 48 &\;\;\; 425\\ &\;\;\; 384\\ \hline &\;\;\;41\\ \hline \end{array}$$

So we have got remainder as 41.So we can say that 41 is the smallest required number to be subtracted from 825 to get a perfect square.
Therefore, we have new number as :825-41=784

So, $$\sqrt{784}=28$$

(v)
$$\qquad \begin{array}{c|lcr} &63\\ \hline 6 &\;\; \;\bar{40}\bar{00}\\ &-36\\ \hline 123 &\;\;\; 400\\ &\;\;\; 369\\ \hline &\;\;\;31\\ \hline \end{array}$$

So we have got remainder as 31.So we can say that 31 is the smallest required number to be subtracted from 4000 to get a perfect square.
Therefore, we have new number as :4000-31=3969

So, $$\sqrt{3969}=63$$