Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.4 National olympiad Prove it South Africa

At the start of the Mighty Mathematicians Football Team's first game of the season, their coach noticed that the jersey numbers of the 22 players on the field (11 players per team) were all the numbers from $1$ to $22$ . At half-time, the coach substituted her goal-keeper (who had the number $1$ on her jersey) for a reserve player. The coach then noticed that after the substitution, no two players on the field had the same jersey number and that the sum of the jersey numbers of each of the teams were exactly equal.

a) What is the smallest (positive) possible jersey number of the reserve player?

b) What is the greatest (positive) possible jersey number of the reserve player?

[10]

Solution

The sum from $2$ to $22$ is $2 + 3 + \cdots + 22 = \frac{22(23)}{2} - 1 = 11 \times 23 - 1 = 252$ .

Therefore the new number must be even or otherwise the sum can't be exactly divisible by $2$ .

The smallest possible even number we can use is $24$ . To see that this is indeed possible take
$24 + 22 + 21 + 20 + 19 + 12 + 6 + 5 + 4 + 3 + 2 = 138 = 18 + 17 + 16 + 15 + 14 + 13 + 11 + 10 + 9 + 8 + 7.$

To find the greatest number possible, we try and put the $10$ smallest numbers with the greatest number. The sum of the $11$ numbers from $12$ to $22$ is
$12 + 13 + \cdots + 22 = \frac{22(23)}{2} - \frac{11(12)}{2} = 11 \times 23 - 11 \times 6 = 11 \times 17 = 187.$

Hence the largest total can be $187$ . Subtracting the numbers $2$ to $11$ from this gives us
$187 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 = 122,$
which is an even number. Thus $122$ is the largest possible jersey number to achieve this.

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