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

(1) 21 (2) 25

^{ }(3) 41 (4) 67^{ }(5) 73
Solution follows here:

__Solution:__
It’s a good problem on numbers.

Minimum possible sum of 4- consecutive two-digit odd numbers
= 11+13+15+17 =

**56**
Maximum possible sum of 4- consecutive two-digit odd numbers
= 93+95+97+99 =

**384**
-----(1)

Let the four numbers be n,n+2,n+4,n+6 => Sum = 4n+12

Given that, (4n+12)/10 is a perfect square, let it be x

^{2}
=> 4n+12 = 10x

^{2 }---(2)
56 < 10x

^{2}< 384 => 6 < x^{2}< 38 => x^{2}= 9 or 16 or 25 or 36 ---(3)
Form (2), 10x

^{2}is a multiple of 4=> x^{2 }is a multiple of 2 ---(4)
From (3) and (4), the only possible values of x

^{2}are 16 and 36**Case-I:**

x

^{2}= 16 => 4n+12 = 10x^{2}= 160 => n= 37 => the odd numbers are:**37,39,41 and 43****Case-II:**

x

^{2}= 36 => 4n+12 = 360 => n= 87 => the odd numbers are:**87,89,91 and 93**
The only possible answer out of all the options is

**41****Answer(3)**
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