The correct answer is option (C):
41
Let's break down this problem step by step to understand why 41 is the correct answer.
We're looking for four consecutive two-digit odd numbers. Let's represent these numbers algebraically. If 'n' is an odd number, then the next three consecutive odd numbers would be n+2, n+4, and n+6.
The sum of these four consecutive odd numbers is: n + (n+2) + (n+4) + (n+6) = 4n + 12
Now, the problem states that when this sum is divided by 10, the result is a perfect square. So, (4n + 12) / 10 = a perfect square. Let's call this perfect square 'k^2'. Therefore:
(4n + 12) / 10 = k^2
4n + 12 = 10k^2
4n = 10k^2 - 12
n = (10k^2 - 12) / 4
n = (5k^2 - 6) / 2
Since 'n' represents an odd number, and we're looking for two-digit odd numbers, we know that the expression (5k^2 - 6) / 2 must result in a two-digit odd number. We also know that k must be an integer, otherwise k^2 won't be an integer and the entire expression wouldn't result in an integer (and hence an odd number). Let's test different integer values of k and determine if the result satisfies the conditions. We're looking for when n, n+2, n+4, and n+6 are all two-digit numbers.
* If k = 1: n = (5(1)^2 - 6) / 2 = -1/2 (not an integer, thus not applicable)
* If k = 2: n = (5(2)^2 - 6) / 2 = 7 (n+2 = 9, n+4 = 11, n+6 = 13). This gives us 7, 9, 11, 13 as our four numbers. However, these aren't all two-digit numbers.
* If k = 3: n = (5(3)^2 - 6) / 2 = 19. This gives us 19, 21, 23, 25. The sum is 88, and 88/10 = 8.8 (not a perfect square).
* If k = 4: n = (5(4)^2 - 6) / 2 = 37. This gives us 37, 39, 41, 43. The sum is 160. 160/10 = 16, which IS a perfect square (4^2).
We found our answer! The four consecutive odd numbers are 37, 39, 41, and 43, and the condition is met. Of the answer choices, 41 is one of these numbers.