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

1. Funny formation

You are the coach of a football team, whose players have shirt numbers 1 to 11. The goalkeeper wears 1. You must divide the others into defenders, midfielders and forwards.

You would like to arrange your team so that the sum of the shirt numbers in each group (defenders, midfielders, forwards) is divisible by 11?

Give an example, or prove it isn’t possible.

Solution No it isn’t possible.

The sum of numbers from 1 to 11 is 66. So the shirt number total for the outfield players is 66- 1 = 65.

If the sums of shirt numbers of defenders, midfielders and forwards are all divisible by 11, the so is the sum of shirt numbers of all these three groups together. But we know this is false, since 11 does not divide 65.

2. Pals or not

When we first learn our times tables, the 11-times table feels delightfully simple:

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

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

If we carry on, 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 multiplying by 11 works. When a two-digit number has digits a and b, the product with 11 is formed by writing down the first digit, then their sum, then the second digit — provided no carrying is needed.

i.e. 11 × 52 = 572,

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

i) matching digits (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 works only while the middle digit stays below 10, which limits us to these four cases.

ii) “staircase” numbers (four solutions)

Now look at numbers where the second digit is one larger 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 final case (one solution)

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

3. Big divide

Less well known than other divisibility rules, there is a simple way to test for 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 (including 0), 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 digits 0–9 exactly once, make the largest possible 10-digit number that is divisible by 11.

**Solution **9876524130​

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

Using the divisibility test for 11, we compare the sum of the digits in odd positions with the sum 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, not a multiple of 11.

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

To keep the number as large as possible, try to preserve the descending prefix. Suppose the number began

987654 ⋯

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

Now try keeping the prefix

98765 ⋯

These digits contribute 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 large as possible, put the largest remaining digits, 4 and 3, into odd positions, and 2,1,0 into 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 the odd-position sum is 28, the even-position 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 Maths Schools for these puzzles. The UK has eleven of these schools, each attached to a university, which are state state sixth forms for 16–19 year olds who love maths. To find out more about their website is umaths.ac.uk.

I’ve been setting a puzzle here on alternate Mondays since 2015. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.