PK 2018 Q15
15. How many 3-digit numbers are there with the property that the 2-digit number obtained by deleting the middle digit is equal to one ninth of the original 3-digit number?
- A 1
- B 2
- C 3
- D 4
- E 5
Pink Kangaroo 2018, Question 15. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D 4
Let the three digits be \(a\), \(b\), \(c\). Then the value of the 3-digit number is \(100a + 10b + c\) and the 2-digit number (made by deleting the middle digit) is \(10a + c\).
Hence \(100a + 10b + c = 9(10a + c)\), which gives \(10(a + b) = 8c\), or \(5(a + b) = 4c\). Hence \(c\) is divisible by 5. But \(c\) cannot be zero (because this would mean \(a + b = 0\), and the digits cannot all be zero). So \(c = 5\). Then \(a + b = 4\), giving four possibilities: \(a = 1\), \(b = 3\) or \(a = 2\), \(b = 2\) or \(a = 3\), \(b = 1\) or \(a = 4\), \(b = 0\). These all give valid solutions (135 = 9 × 15, 225 = 9 × 25, 315 = 9 × 35, 405 = 9 × 45).
Solution © UK Mathematics Trust, from the PK 2018 solutions.