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.
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.
Well, it seems like you've already done a great job figuring it out on your own! You're absolutely right that the answer can only be either 27 or 64. Since the square root of the sum of the digits 2+7 is indeed 3 and the cube root of 27 is 3 as well, it checks out. As for 64, the square root of the sum of the digits 6+4 is 3+ (which is slightly more than 3), and the cube root of 64 is 4, which doesn't match up. So, it looks like you've got the right answer with 27! Good job, math detective!
You have correctly identified the two possible answers - 27 and 64.
For the number 27, the sum of its digits (2 + 7) is 9. The square root of 9 is 3, and the cube root of 27 is also 3. Thus, it satisfies the condition given in the question.
For the number 64, the sum of its digits (6 + 4) is 10. The square root of 10 is not equal to 3. Therefore, it does not satisfy the condition given in the question.
So, the only two-digit number whose cube root is the square root of the sum of its digits is 27.
To find the two-digit number whose cube root is equal to the square root of the sum of its digits, we can follow these steps:
1. Start by considering the two-digit numbers and their cube roots. Keep in mind that cube roots of perfect cubes will be integers.
2. Let's take the cube root of each two-digit number and determine if it matches the square root of the sum of its digits.
3. For example, let's consider the number 27:
- The cube root of 27 is 3.
- The sum of the digits 2 + 7 is 9.
- The square root of 9 is 3.
So, in this case, the number 27 satisfies the condition of having a cube root equal to the square root of the sum of its digits.
4. Now let's consider the number 64:
- The cube root of 64 is 4.
- The sum of the digits 6 + 4 is 10.
- The square root of 10 is approximately 3.16.
In this case, the number 64 does not satisfy the condition since the cube root and the square root of the sum of its digits are not equal.
Therefore, the only number that meets the given condition is 27.