Question

In a five-digit number 1b6a3, then a is the
greatest single-digit perfect cube and twice of
it exceeds b by 7. Then the sum of the number
and its cube root is​

Answer 1

Answer:

19710

Step-by-step explanation:

2³ = 8

3³ = 27 which is not single digit thus we select a = 8 here.

2*a = 16, 7 less than 16 = 9. thus b = 9.

Thus number becomes 19683.

Cube root of 19683 is 27.

Sum = 19710