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Earlier today, we presented three puzzles involving the digit 11. Below, presented once more complete with their solutions.

1. The Team Formation

Imagine you are the manager of a soccer team. The players have shirt jersey numbers from 1 to 11, with the goalkeeper wearing number 1. The challenge is to divide the other ten outfield players into three groups: defenders, midfielders, and forwards.

The question is: Is it possible to set up the team so that the sum of the shirt numbers in every unit is evenly divided by 11?

Give one arrangement, or prove it isn’t possible.

Solution

This arrangement is impossible.

The sum total of all numbers from 1 to 11 is 66. So, the combined total for the ten field players is 66 - 1 = 65.

If the sums for the defenders, midfielders, and forwards were each divisible by 11, then their combined sum would also be a multiple of 11. But, we know this is false, since 65 is not divisible by 11.

2. Palindromic Products

As we learn our times tables, the 11-times table feels delightfully simple.

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

Every answer are palindromes (digits that read the same forwards and backwards).

If we continue up to 11 × 99, how many more answers are palindromic?

Solution

We find an additional nine.

Consider the method of multiplying by 11. For a two-digit figure has digits a and b, the result with 11 is created by writing the first digit, then their sum, then the second digit—provided no carrying is needed.

For example, 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 the product with 11 produces 121, 242, 363, and 484—each one palindromic.

This holds true only as long as the middle digit stays below 10, which limits us to these four instances.

ii) “Staircase” Numbers (four cases)

Now consider numbers where the last digit is the successor of the first. Here, the outer digits of the product are the same, again giving palindromes.

• 11 × 56 = 616
• 11 × 67 = 737
• 11 × 78 = 858
• 11 × 89 = 979

iii) The Final Case (single case)

Go a little higher, the product becomes four digits. The leading digit will be a 1, so the only hope for a palindrome is when the last digit is also 1. Testing 91 reveals it succeeds! 11 × 91 = 1001.

3. Big Divide

Not as famous than other divisibility rules, there is a simple method to check 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.

As an example, for 132 we calculate +1 - 3 + 2 = 0, so 132 is divisible by 11.

Employing each of the digits 0–9 only once, create the greatest 10-digit number that is a multiple of 11.

Answer

The number is 9876524130.

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

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

• Digits in odd places: 9, 7, 5, 3, 1 total 25.
• Even positions: 8, 6, 4, 2, 0 sum to 20.

The difference is 5, which is not divisible by 11.

Because the sum of all digits 0–9 is 45, the two sums must invariably total 45, so their difference can never be 0. The closest multiple of 11 we can aim for is therefore 11.

To keep the number as high as possible, we try to preserve the descending prefix. If the number began 987654…

These initial digits contribute a difference of (9+7+5) - (8+6+4) = 3. The other digits 3,2,1,0 would need to contribute 8, but in the best arrangement they can only contribute (3+1) - (2+0) = 2. So no number starting 987654… will succeed.

Next, attempt keeping the prefix 98765…

These digits give a difference of (9+7+5) - (8+6) = 7. The other digits 4,3,2,1,0 must contribute exactly 4.

To make as large as possible the difference, put the largest remaining digits, 4 and 3, into odd places, and 2,1,0 into even positions. This yields (4+3) - (2+1+0) = 4, exactly what we need.

Ordered to maintain the number as large as possible, this gives us 9876524130.

A quick check confirms: the sum in odd places is 28, the sum in even places is 17, and 28 - 17 = 11, so the number is indeed a multiple of 11.

We hope you enjoyed these challenges. More puzzles are on the way soon.