Question

Find the square roots of \(100\) and \(169\) by the method of repeated subtraction.

Answer 1

We know that the sum of the first n odd natural numbers is \(n^2\).
Consider \(\sqrt{100}\).
(i) \(100 - 1 = 99\) (ii) \(99 - 3 = 96\) (iii) \(96 - 5 = 91\)
(iv) \(91 - 7 = 84\) (v) \(84 - 9 = 75\) (vi) \(75 - 11= 64\)
(vii) \(64 - 13 = 51\) (viii) \(51 - 15 = 36\) (ix) \(36 - 17 = 19\)
(x) \(19 - 19 = 0\)
We have subtracted successive odd numbers starting from \(1\) to \(100\), and obtained \(0\) at \(10^{th}\) step.
Therefore, \(\sqrt{100}=10\)

The square root of \(169\) can be obtained by the method of repeated subtraction as follows.
(i) \(169 - 1 = 168\) (ii) \(168 - 3 = 165\) (iii) \(165 - 5 = 160\)
(iv) \(160 - 7 = 153\) (v) \(153 - 9 = 144\) (vi) \(144 - 11 = 133\)
(vii) \(133 - 13 = 120\) (viii) \(120 - 15 = 105\) (ix) \(105 - 17 = 88\)
(x) \(88 - 19 = 69\) (xi) \(69 - 21 = 48\) (xii) \(48 - 23 = 25\)
(xiii)\(25 - 25 = 0\)
We have subtracted successive odd numbers starting from \(1\) to \(169\), and obtained \(0\) at \(13^{th}\) step.
Therefore, \(\sqrt{169} = 13\)