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# Find the number of digits in the square roots of each of the following numbers:a. 64b. 144c. 4489d. 27225e. 390625

Last updated date: 28th Mar 2023
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Hint: Count the number of digits and for even $\left( \dfrac{n}{2} \right)$and for odd $\left( \dfrac{n+1}{2} \right)$and we will get the number of digits in the square root of the numbers.

(a) 64
The number of digits in 64 = n = 2.
Here, n=2 is even
$\therefore$Number of digits in square root = $\dfrac{n}{2}=\dfrac{2}{2}=1$

(b) 144
Number of digits in 144 = n = 3
Here, n = 3 is odd
$\therefore$Number of digits in square root = $\dfrac{n+1}{2}=\dfrac{3+1}{2}=2$

(c) 4489
Number of digits in 4489 = n = 4
Here, n=4 is even
$\therefore$Number of digits in square root =$\dfrac{n}{2}=\dfrac{4}{2}=2$

(d) 27225
Number of digits in 27225 = n = 5
Here, n = 5 is odd
$\therefore$Number of digits in square root =$\dfrac{n+1}{2}=\dfrac{5+1}{2}=\dfrac{6}{2}=3$

(e) 390625
Number of digits in 390625 = n = 6
Here, n=6 is even
$\therefore$Number of digits in square root =$\dfrac{n}{2}=\dfrac{6}{2}=3$

Note:
(i) $\sqrt{64}=8$, number of digits in square root = 1.
(ii) $\sqrt{114}=12$, number of digits in square root = 2.
(iii) $\sqrt{4489}=67$, number of digits in square root = 2.
(iv) $\sqrt{27225}=165$, number of digits in square root = 3.
(v) $\sqrt{390625}=625$, number of digits in square root = 3.