Consider the following multiplication problem: (PQ)×3 = RQQ, where P, Q and R are different digits and R≠0.…

Consider the following multiplication problem: (PQ)×3 = RQQ, where P, Q and R are different digits and R≠0. What is the value of (P+R)÷Q?
  1. 1
  2. 2
  3. 5
  4. Cannot be determined due to insufficient data

Solution

PQ × 3 = RQQ means $(10P + Q) \times 3 = 100R + 10Q + Q$, so $30P + 3Q = 100R + 11Q$, i.e. $30P = 100R + 8Q$. Since 30P and 100R end in 0, 8Q must end in 0, giving Q = 5. Then $30P = 100R + 40$, so $3P = 10R + 4$. R = 2 gives P = 8. Thus 85 × 3 = 255, so P = 8, Q = 5, R = 2. $(P + R)/Q = (8 + 2)/5 = 2$.

Asked in: CSAT 2021

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