The highest possible value of date and month in a calendar are 31 and 12 respectively.
∴ 31d + 12m = 487 …(i)
Since there is only one equation and number of variables is two, we cannot find the solution of this equation by normal method.
But, there is other information given in the question implicitly. The date can be any of the integers 1, 2, …, 31 and the month can be any of the integers 1, 2, …, 12. By combining this information with equation (i) we can find a unique solution.
Consider equation (i), when left hand side and right hand side of this equation is divided by 12, the remainders should be same. [Note: we are taking 12 here, because 12 is the smaller coefficient]
Second term 12m will not give any remainder when divided by 12, and the remainder when 31d is divided by 12 will be same as the remainder when 7d is divided by 12. Also, the remainder when 487 is divided by 12 will be 7.
∴ From L.H.S. (the remainder when 7d is divided by 12) = from R.H.S. (the remainder when 487 is divided by 12)
∴ The remainder when 7d is divided by 12 = 7
To satisfy the above condition the minimum value of d could be 1. [Because when 7 is divided by 12 the remainder is 7 itself]
The corresponding value of m can be obtained by substituting d = 1 in equation (i).
∴ When d = 1, m = 38
We have to discard this pair of (d, m), because the month of birth cannot be 38.
To find the other pair of values for d and m, we increase the value of d by 12 (the coefficient of m) and decrease the value of m by 31 (the coefficient of d).
∴ The other pair for (d, m) will be (13, 7), (25, −24) and so on. Here also, we have to reject the second pair because month cannot be negative. Therefore no other pair is possible, and the only valid pair of date and month possible is d = 13 and m = 7.
∴ My complete date of birth is 13/7/1976.
∴ Sum of my date and month of birth = 13 + 7 = 20
Hence, option 5.
