Consider the following addition problem : 3P+4P+PP+PP = RQ2; where P, Q and R are different digits. What is the arithmetic mean of all such possible sums?
102
120
202
220
3P + 4P + PP + PP = RQ2 Or 30 + P + 40 + P + 10P + P + 10P + P = 100R + 10 Q + 2 Or 24P + 70 = 100R + 10 Q + 2 Or 20P + 70 + 4P = 100R + 10 Q + 2 The unit digit of the resultant is 2. It will be obtained when 4 is multiplied by P. So, P must be 3, or 8. If P = 3, then: 24P + 70 = 24 × 3 + 70 = 72 + 70 = 142 If P = 8, then: 24P + 70 = 24 × 8 + 70 = 192 + 70 = 262 Arithmetic sum of 142 and 262 = (142 + 262)/2 = 202
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