The cube of a two-digit number may have seven or more digits. (True or False)

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Solution

For instance
${(10)}^{3} = 10 \times 10 \times 10 = 1000$ which is a $4$ digit number
${(45)}^{3} = 45 \times 45 \times 45 = 91125$ which is a $5$ digit number
The cube of a two-digit number can not have seven or more digits
Therefore,The given statement is False

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Similar questions

State true or false.

(i) Cube of any odd number is even.

(ii) A perfect cube does not end with two zeroes.

(iii) If square of a number ends with 5, then its cube ends with 25.

(iv) There is no perfect cube which ends with 8.

(v) The cube of a two digit number may be a three digit number.

(vi) The cube of a two digit number may have seven or more digits.

(vii) The cube of a single digit number may be a single digit number.

Question 2(v)

State True or False.
The cube of a two digit number may be a three digit number.

Question 76

State whether the following statement is true or false:

The cube of a one-digit number cannot be a two-digit number.

Question 2(vi)

State True or False.
The cube of a two digit number may have seven or more digits.

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