PK 2019 Q21
21. Each of the positive integers \(a\), \(b\), and \(c\) has three digits, and for each of these integers the first digit is the same as its last digit. Also \(b = 2a + 1\) and \(c = 2b + 1\). How many possibilities are there for the integer \(a\)?
- A 0
- B 1
- C 2
- D 3
- E more than 3
Pink Kangaroo 2019, Question 21. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 2
Let the digits of \(a\) be ‘\(pqp\)’, so \(a = 101p + 10q\).Also \(c = 2b + 1 = 2(2a + 1) + 1 = 4a + 3\) which is \(404p + 40q + 3\). This is less than 1000, so \(p = 1\) or 2.
If \(p = 2\), then \(c = 808 + 40q + 3\). This ends with digit 1 (8 + 3 = 11) but \(c\) is a 3-digit number greater than 808 so can’t begin with 1.
If \(p = 1\), then \(c = 404 + 40q + 3\). This ends with digit 4 + 3 = 7, so must also begin with 7, hence \(700 \leq 404 + 40q + 3 < 800\) and thus \(293 \leq 40q < 393\). Therefore \(q = 8\) or 9.
When \(q = 8\), this gives \(a = 181\), \(b = 363\), \(c = 727\). When \(q = 9\), this gives \(a = 191\), \(b = 383\), \(c = 767\).
Hence there are two possibilities for \(a\), 181 and 191.
Solution © UK Mathematics Trust, from the PK 2019 solutions.