Math, asked by aayushkumarjaiswal20, 2 months ago

3. For an n-digit number where n is odd, the number of digits in its square root are​

Answers

Answered by ankitamikku
0

Answer:

no of digits in a perfect square is n

If n is odd then no of digits in its square roots is  2n+1

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Step-by-step explanation:

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