Did you solve it? The numbers all go to 11 | Mathematics

Earlier today I asked you these three problems regarding the number 11. Here they are again with solutions.

1. Fun training

You are the coach of a football team whose players wear jerseys numbered 1 to 11. The goalkeeper wears 1. You must divide the others into defenders, midfielders and attackers.

Do you want to organize your team so that the sum of the jersey numbers of each group (defenders, midfielders, attackers) is divisible by 11?

Give an example or prove that this is not possible.

Solution No, that’s not possible.

The sum of the numbers 1 to 11 is 66. The total of the outfield players’ jersey numbers is therefore 66 – 1 = 65.

If the sum of the jersey numbers of defenders, midfielders and attackers is divisible by 11, the same is true for the sum of the jersey numbers of these three groups combined. But we know this is wrong since 11 does not divide 65.

2. Friends or not

When we first learn our multiplication tables, the 11 table seems delightfully simple:

11 × 1 = 11
11 × 2 = 22
11 × 3 = 33
…
11 × 9 = 99

All answers are palindromes (numbers that read the same backwards and forwards).

If we continue, up to 11 x 99, how many more answers are palindromes?

[Hint. At least one! For example, 11 × 56 = 616.]

Solution nine more

Consider how multiplication by 11 works. When a two-digit number has digits A And bthe product with 11 is formed by writing the first digit, then their sum, then the second digit — provided no carryover is necessary.

i.e. 11 × 52 = 572,

since the middle number is simply 5+2=7.

i) matching numbers (four solutions)

If the two digits are the same, as in 11, 22, 33 or 44, then multiplying by 11 produces 121, 242, 363 and 484 – all palindromes.

This only works as long as the middle number remains below 10, which limits us to these four cases.

ii) “Staircase” numbers (four solutions)

Now let’s look at numbers whose second digit is greater than the first. In these cases, the outer digits of the product match, again giving palindromes:

11 × 56 = 616

11 × 67 = 737

11 × 78 = 858

11 × 89 = 979

iii) The last case (a solution)

Go a little higher and the product is four figures. The first number will be a 1 so the only hope of getting a palindrome is when the last number is a 1, so let’s test 91. Surprisingly, it works! 11×91=1001.

3. Great divide

Less known than the other divisibility rules, there is a simple way to test divisibility by 11.

Take the digits of a number and add them alternately with plus and minus signs (starting with a plus). If the result is a multiple of 11 (0 inclusive), then the original number is divisible by 11.

For example, for 132 we get +1-3+2 = 0, so 132 is divisible by 11.

Using each of the numbers 0 through 9 exactly once, create the largest possible 10-digit number divisible by 11.

Solution 9876524130​

The largest 10-digit number using the digits 0 through 9 once each is 9876543210.

Using the divisibility test for 11, we compare the sum of digits in odd positions with the sum of digits in even positions. For this number,

• odd positions: 9,7,5,3,1 sum to 25;

• even positions: 8,6,4,2,0 sum to 20.

The difference is 5 and not a multiple of 11.

Since the total of the digits 0 to 9 is 45, the two sums must always add up to 45, so their difference can never be 0. So the closest multiple of 11 we can aim for is 11.

To make the number as large as possible, try to keep the prefix descending. Suppose the number begins

987654 ⋯

The figures set so far contribute to a difference in
(9+7+5)−(8+6+4)=3.
So the remaining digits 3,2,1,0 should contribute to 8. But even arranged in the best possible way, they can only contribute (3+1)−(2+0)=2.
So no number starting with 987654… can work.

Now try to keep the prefix

98765 ⋯

These numbers contribute to a difference of (9+7+5)−(8+6)=7.
So the remaining digits 4,3,2,1,0 must contribute exactly 4.

To make the difference as big as possible, place the largest remaining numbers, 4 and 3, in odd positions and 2,1,0 in even positions. This gives
(4+3)−(2+1+0)=4, exactly what we need.

Arranged to keep the number as large as possible, this gives 9876524130​.

A quick check confirms that the odd sum is 28, the even sum is 17, and 28−17=11, so the number is divisible by 11.

I hope you had fun. I’ll be back in two weeks.

Thanks to the University Schools of Mathematics for these puzzles. The UK has eleven such schools, each attached to a university, which are state sixth form schools for young people aged 16 to 19 who enjoy mathematics. To find out more on their website, visit umaths.ac.uk.

I’ve been putting up a puzzle here every other Monday since 2015. I’m always looking for great puzzles. If you want to suggest one, send me an email.