IMC 2012 Q8
8. Seb has been challenged to place the numbers 1 to 9 inclusive in the nine regions formed by the Olympic rings so that there is exactly one number in each region and the sum of the numbers in each ring is 11. The diagram shows part of his solution.
What number goes in the region marked * ?
- A 6
- B 4
- C 3
- D 2
- E 1
Intermediate Mathematical Challenge 2012, Question 8. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: A 6
Referring to the diagram, \(a = 11 - 9 = 2\);
\(b = 11 - 5 - a = 4\); \(f = 11 - 8 = 3\). So the values of \(c\), \(d\) and \(e\) are 1, 6, 7 in some order.
We need to have \(b + c + d = 11\) and \(d + e + f = 11\). Therefore, as \(b = 4\) and \(f = 3\), \(c + d = 7\) and \(d + e = 8\). The only solution with \(c\), \(d\), \(e\) chosen from 1, 6, 7 is \(c = 6\), \(d = 1\) and \(e = 7\). So 6 must be in the region labelled *.
Solution © UK Mathematics Trust, from the IMC 2012 solutions. Longer versions with investigations: Solutions and investigations (PDF).