Question

The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?

• A. 21
• B. 25
• C. 41
• D. 67
• E. 73

Hint:

Always use the average for consecutive numbers to simplify the sum conditions.

Answer 1

The Correct Option is C

Let the numbers be $n-3, n-1, n+1, n+3$. Their sum: \[ 4n \] Condition: $\frac{4n}{10}$ is a perfect square $\Rightarrow \frac{2n}{5} = k^2$ for some integer $k$. Thus: \[ n = \frac{5k^2}{2} \] $n$ must be integer ⇒ $k^2$ even ⇒ $k$ even ⇒ $k=2m$: \[ n = \frac{5 \cdot 4m^2}{2} = 10m^2 \] Possible $n$: 10, 40, 90, etc. Among choices, only 41 is close to form of $n \pm 1, n \pm 3$ from $n=40$. So 41 is valid. \[ \boxed{41} \]