Question

Find a three digit number which, when reversed, becomes equal to 17 times the square of its cube root. ​

Answer 1

Answer:

Assuming the number is of the form 'xyz', we can write it mathematically as 100x+10y+z and the number with digits in reversed order would be 100z+10y+x

Hence the question is essentially to solve for

100z+10y+x = 17 * (100x+10y+z)^2/3

Clearly not that easy.

But if we try to think smartly we can solve this question easily.

1. The RHS contains cube root of a 3 digitn number. So list all the 3 digit numbers which are perfect cubes:

5^3 = 125

6^3 = 216

7^3 = 343

8^3 = 512 &

9^3 = 729

2. Now check if 17 * square of the cube root is the reversed number

17 * 5^2 = 425

17 * 6^2 = 612

17 * 7^2 = 833

17 * 8^2 = not a 3 digit no

17 * 9^2 = not a 3 digit no

3. As we can clearly see only 612 or 216 satisfies.

Hence the answer is 216.