UPSC CSE Prelims 2021
The given problem is an alphanumeric addition: 3P + 4P + PP + PP = RQ2, where P, Q, and R are distinct single digits.
To solve this, we first expand the terms based on their place values:
Substituting these expanded forms into the original equation:
(30 + P) + (40 + P) + (11P) + (11P) = 100R + 10Q + 2
Combining like terms on the left side:
70 + 24P = 100R + 10Q + 2
The unit digit of the resultant sum (RQ2) is specified as 2. This implies that the unit digit of the expression (70 + 24P) must also be 2. Since the unit digit of 70 is 0, the unit digit of 24P must be 2. This condition is met when the unit digit of the product 4 × P is 2.
The possible single-digit values for P that satisfy this condition are:
Next, we calculate the possible sums (RQ2) for each valid value of P:
Case 1: P = 3
Substituting P = 3 into the simplified equation: 70 + 24 × 3 = 70 + 72 = 142.
Here, the sum is 142. Comparing with RQ2, we have R=1, Q=4, and P=3. All these digits (1, 4, 3) are distinct, fulfilling the problem's condition. Thus, 142 is a valid sum.
Case 2: P = 8
Substituting P = 8 into the simplified equation: 70 + 24 × 8 = 70 + 192 = 262.
Here, the sum is 262. Comparing with RQ2, we have R=2, Q=6, and P=8. All these digits (2, 6, 8) are distinct, fulfilling the problem's condition. Thus, 262 is a valid sum.
The possible sums are 142 and 262.
The arithmetic mean of these possible sums is calculated as:
Arithmetic Mean = (142 + 262) / 2 = 404 / 2 = 202.
Options A (102), B (120), and D (220) are incorrect because they do not correspond to the arithmetic mean of the valid sums derived from the given alphanumeric puzzle. The only two valid sums are 142 and 262, and their arithmetic mean is 202.