# Find the number of digits in the square roots of each of the following numbers:

a. 64

b. 144

c. 4489

d. 27225

e. 390625

Last updated date: 28th Mar 2023

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Answer

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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.

Complete step-by-step answer:

(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.

Complete step-by-step answer:

(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.

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