The members of the Much Smashing tennis club were relaxing in the pub after a strenuous afternoon on the courts.
“How did it go?” Basil asked Herbert.
“Not too bad,” Herbert said. “I played one set with Onslow and beat him 6-3. But we couldn’t get our serves working properly. Five games went against serve.”
“Room for improvement,” said Basil. ‘Did you serve first, or was it Onslow?”
Who served first?
Prospect invites you to solve the puzzle and send us the solution. Correct answers will be entered into a draw. The winner receives A Brief Guide to the Great Equations by Robert P Crease (Constable & Robinson) Send your solution to? email@example.com by 17th April. The winner will be announced in our May issue. Last month’s winner was Daniel Bjelis, Durham. Last month’s answer The largest multiple of 11 that uses each digit exactly once is 9,876,524,130. Let the sum of the “odd” digits—positions 1, 3, 5 and so on from the left—be a. Let the sum of the “even” digits—positions 2, 4 6 and so on—be b. We want a–b to be a multiple of 11. And we know that a+b = 45, as it is the sum of all ten digits. If a–b is even then a+b would also be even, but 45 is odd. So a–b is an odd multiple of 11. Now, a ? 9+8+7+6+5 = 35, and a ? 0+1+2+3+4 = 10. The same goes for b. Therefore a–b must be either 11 or -11. We can quickly rule out 9,876,543,210 and 9,876,543,201. If the solution were to start 9,876,543 then a must be at least 24 and b must be at least 18. The remaining digits only add up to 3; whichever way they are placed, the difference between a and b is less than 11. For similar reasons, numbers starting 9,876,54 are ruled out. Next we try numbers starting 9,876,5. None of the numbers starting 9,876,53 work. Turning to 9,876,52, a (so far) is 21 and b is 16, which differ by 5. The remaining digits need to increase this difference by 6—but we can achieve this by using 4 and 3 in the “odd” positions and 1 and 0 in the “even” positions. To get the biggest number, we place the largest digits of each pair first, giving us 9,876,524,130.