Enigmas & puzzles

The rule of eleven
March 1, 2009

"Did you know," said Innumeratus, "that the number 618,235,079 contains all of the digits from 0 to 9, except the number 4, and is exactly divisible by 11?"

Mathophila did a quick sum in her head. "You're right, it's 11×56,203,189. I wonder if you can find a number that uses all ten digits which is divisible by 11?"

A moment later: "Yes, you can."

"You can't have tried all the possible arrangements in that short a time!"

"No, don't be silly. I used an old trick. Take your number, for instance. Add up every other digit, starting from the left: 6+8+3+0+9 = 26. Now add up the remaining digits: 1+2+5+7 = 15. The difference is 11, and since that is divisible by 11, then so is your number. If it wasn't divisible by 11, your number wouldn't be either. The rule works for any whole number.

"So I just checked if I could split the digits 0-9 into two sets of five so that their sums differ by a multiple of 11. And I could: 0+4+7+8+9 = 28 and 1+2+3+5+6 = 17. So one example is 1,024,375,869. And I could shuffle the order of the two sets, so there must be lots of ways to do it."

Innumeratus gave her a look. "What's the largest number like that, I wonder?"

What is the largest multiple of 11 that uses each digit from 0 to 9 exactly once?



Prospect invites you to solve the puzzle and send us the solution. Correct answers will be entered into a draw. The winner will receive a copy of "e: The Story of a Number by Eli Maor" (Princeton University Press, £9.50). Send your solution to answer@prospect-magazine.co.uk by 13th March.

The winner will be announced in our April issue. Last month's winner was Will Broad, London SW12.

Last month's answer

The largest circle has radius 7. The other two have radii 5 and 1.

Suppose the circles have radii x, y, z where x is the smallest and z the largest. All three are integers. The areas are ?x2, ?y2 and ?z2. The blue region has area ?(z2–y2) and the yellow ?(y2–x2), and these must be equal. So z2–y2 = y2–x2 and therefore x2+z2 = 2y2.

We know that all three numbers are less than or equal to ten. The only solution is this range is x = 1, y = 5, z = 7.