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?
Explanation:
The four consecutive two-digit odd numbers will have (1, 3, 5, 7) or (3, 5, 7, 9) or (5, 7, 9, 1) or (7, 9, 1, 3) or (9, 1, 3, 5) as units digits.
As the sum divided by 10 yields a perfect square, the sum is a multiple of 10.
∴ The units digits have to be (7, 9, 1, 3).
Thus the four numbers will be (10x + 7), (10x + 9), (10x + 11) and (10x + 13),
where 0 < x < 9 (as each of these numbers is a two digit number)
Sum of these numbers = 40x + 40 = 40(x + 1)
Now, 40(x + 1)/10 = 4(x + 1) is a perfect square
As 4 is a perfect square, (x + 1) is some perfect square < 10
If x + 1 = 4, x = 3, and the four numbers are 37, 39, 41 and 43
If x + 1 = 9, x = 8, and the four numbers are 87, 89, 91 and 93
Hence, option (c).