Question

शो द 1331 इज नॉट अ परफेक्ट स्क्वायर​

Answer 1

1331=we divide 1331 into groups of three digits starting from the right. So 1331 has two groups one is 331 and another is 1.

For first group 331, the digit 1 is at one's place .1 comes at a unit place of a number only when its cube root ends with 1. So one's place of the required cube root is 1.

For another group, i.e.1 1

3

=1 and 2

3

=8 so 1 lies between 0 and 8.the smaller number among 1 and 2 are 1. So the one's place of 1 is 1 and ten's place of cube root 1331 is 1

Hence

3

1331

=11

For 4913

we divide 4913 into groups of three-digit starting from the right. So 4913 has two groups one is 913 and another is 4.

For first group 913, the digit 3 is at one's place .3 comes at a unit place of a number only when its cube root ends in 7. So one's place of the required cube root is 7.

For another group, i.e.1 1

3

=1 and 2

3

=8 so 4 lies between 1 and 8.the smaller number among 1 and 2 are 1. So the one's place of 1 is 1 and ten's place of cube root 4913 is 1

Hence

3

4913

=17

For 12167

we divide 12167 into groups of three-digit starting from the right. So 12167 has two groups one is 167 and another is 12.

For first group 167, the digit 7 is at one's place .7 comes at a unit place of a number only when its cube root ends in 3. So one's place of the required cube root is 3

For another group, i.e.12 2

3

=8 and 3

3

=27 so 12 lies between 8 and 27.the smaller number among 2 and 3 are 2. So the one's place of 2 is 2itself and ten's place of cube root 12167 is 2

Hence

3

12167

=23

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