## Friday, 4 November 2011

### 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