What is the least integer whose square root is an irrational number between 5 and 7?
A. 13
B. 6
C. 25
D. 26
To solve this problem, we need to find the least possible integer $n$ such that $\sqrt{n}$ is between 5 and 7. To find the least possible $n$, we can start by finding the square of 5 and the square of 6, and then keep incrementing until we find a square greater than 49 (since 7 squared equals 49).
Squaring 5, we get $5^2 = 25$.
Squaring 6, we get $6^2 = 36$.
Let's calculate the square of 7 to see if it is less than or greater than 49.
Squaring 7, we get $7^2 = 49$.
Since 49 is the smallest square greater than 49, the least possible integer $n$ whose square root is an irrational number between 5 and 7 is $\boxed{\textbf{(D) } 26}$.
To find the least integer whose square root is an irrational number between 5 and 7, we need to find the smallest perfect square between 5^2 and 7^2.
5^2 = 25
7^2 = 49
The smallest perfect square between 25 and 49 is 26^2 = 676.
The square root of 676 is 26, which is an integer.
Therefore, the correct answer is D. 26