Numbers-12 (CAT-2006)

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 x²

=> 4n+12 = 10x²                    ---(2)

56 < 10x² < 384 => 6 < x² < 38            => x² = 9 or 16 or 25 or 36      ---(3)

Form (2),  10x² is a multiple of 4=> x² is a multiple of 2        ---(4)

From (3) and (4), the only possible values of x² are 16 and 36

Case-I:

x² = 16 => 4n+12 = 10x² = 160 =>  n= 37 => the odd numbers are: 37,39,41 and 43

Case-II:  

x² = 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)