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, $\sqrt{169} = 13$