Question

the two-digit number whose cube root is the square root of the sum of its digits


THe cube root of 27 is 3.


Only two possible answers; 27 or 64

Looks fairly obvious.


2+7 = 9. Sqrt 9 = 3 and (27)^1/3 = 3. check.

6+ 4 = 10. sqrt 10 = 3+ and (64)^1/3 = 4
doesn't check. or have I missed something.

Answer 1

Let the two-digit number be $ab$, where $a$ is the tens digit and $b$ is the ones digit. Since the cube root of $ab$ is the square root of the sum of its digits, we have
[\sqrt[3]{ab} = \sqrt{a+b}]Squaring both sides, we get
[\sqrt[3]{ab}^2 = \sqrt{a+b}^2]Since the cube of any number is the same as the number itself, we have
[\sqrt[3]{ab}^3 = \sqrt[3]{ab}]This means that
[ab = \sqrt{a+b}]Squaring both sides again, we get
[ab^2 = a+b]Since $ab$ is a two-digit number, we have $b^2 < 10$, which means $a+b < 100$. Therefore, the only possible values for $a+b$ are $1, 4, 9, 16, 25, 36, 49, 64, 81$.

If $a+b = 1$, then $ab$ must be $1$, but $1$ is not a two-digit number.

If $a+b = 4$, then $ab$ must be $2$, but $2$ is not a two-digit number.

If $a+b = 9$, then $ab$ must be $3$, but $3$ is not a two-digit number.

If $a+b = 16$, then $ab$ must be $4$, but $4$ is not a two-digit number.

If $a+b = 25$, then $ab$ must be $5$, but $5$ is not a two-digit number.

If $a+b = 36$, then $ab$ must be $6$, but $6$ is not a two-digit number.

If $a+b = 49$, then $ab$ must be $7$, but $7$ is not a two-digit number.

If $a+b = 64$, then $ab$ must be $8$, but $8$ is not a two-digit number.

If $a+b = 81$, then $ab$ must be $9$, but $9$ is not a two-digit number.

Therefore, the two-digit number whose cube root is the square root of the sum of its digits does not exist.