2016 · Q5
What would be the maximum value of Q in the equation 5P9 + 3R7 + 2QS = 1114, where P, Q, R, S are digits?
Elementary MathematicsNumber System
- A9
- B8
- C5
- D4
Explanation
Units column: 9+7+S ends in 4, so S = 8 with carry 2. Hundreds column: 5+3+2+carry = 11 forces the carry into the hundreds to be 1. Hence the tens column satisfies P+R+Q+2 = 11, i.e. P+R+Q = 9. To maximise Q with P,R ≤ 9, take Q = 9 (then P = R = 0), giving option A.