In the five digit number $1 b 6 a 3$ , a is the greatest single digit perfect cube and twice of it exceeds by $7$ . Then the sum of the number and its cube root is __________.

18700

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11862

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19710

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25320

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Solution

The correct option is B $19710$
$5$ digit no. is $= 1 b 6 a 3$
Greatest single digit perfect cube $= 8$
$∴ a = 8$
Given $b = 2 a - 7 = 2 \times 8 - 7 = 9$
No: $19683$
Ans: $19683 + \sqrt[3]{19683} = 19683 + 27$
$= 19710$

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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.

The product of the digits of a two digit number is 24. If its units's digit exceeds twice its ten's digit by 2; find the number.

2

144

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