Let the no. be 10x + y.No. formed by the interchange of digits = 10y + xWe have y - x = 2 .........(i)y + x = 14 .........(ii)Solving (i) and (ii), we get x = 6, and y = 8∴ the no. is 68.
Let the four numbers be x, x + 2, x + 4 and x + 6.=> x + x + 2 + x + 4 + x + 6 = 36=> 4x + 12 = 36 => x = 6The numbers are 6, 8, 10 and 12.Sum of their squares = 62 + 82 + 102 + 122 = 36 + 64 + 100 + 144 = 344.
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