Hannah was able to solve the initial problem using algebra to find the expressions. Her solution is here.

Irene from King's College of Alicante, and Zach considered what happens when $N$ is odd. Irene described it as:

When $N$ is odd, the number that must be wiped out is $\frac{N+1}{2}$.

Zach had the following description:

N |
Total before wipeout |
Wipeout Number |

3 | 6 | 2 |

5 | 15 | 3 |

7 | 28 | 4 |

9 | 45 | 5 |

11 | 66 | 6 |

We have a formula that the mean is $\frac{t_N - w}{N-1}$, where $t_N$ is the sum of the first $N$ integers and $w$ is the number that is wiped out.

Because $N$ is odd, $N – 1$ must be even. If we substitute this into our formula for the mean , and want to have only positive integers, the numerator must also be even (as odd $\div$ even does not generate a positive integer). This leaves you looking for an even numerator, and comes down to the combination of odd and even pairs in each triangular number.

For example: $t_3 = 1 + 2 + 3$, where $1 + 3$ gives an even number, so $2$ has to be the wipeout number. But $t_9 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$, which gives an odd number, so $5$ is the wipeout number as there are 2 pairs of odd numbers, and two pairs of even numbers that sum together to make an even number divisible by $8$.

Although the wipeout numbers are consecutive, they are actually the median number of each of the sequences ($2$ is the median of $1,2,3$, etc.). The same logic applies: the wipeout number is the remainder.

Suppose $n$ is the missing number. The total is $1+2+3+4+5+6=21$. If there is one less number, then the mean is $\frac{21-n}{5}$.

Substituting values for the missing number $n$:

$n=6$: $21-6=15$. $15 \div 5=3$. *$3$ is less than $3.6$ but fairly close.*

$n=5$: $21-5=16$. $16 \div 5=3.2$. *$3.2$ is a bit closer to $3.6$ so $n<5$*

(Writing the answers to the divisions in decimals, you realise that when dividing increasing numbers by $5$, the outcome increases each time by $0.2$. Knowing this, dividing the next numbers will be easy.)

$n=4$: $21-4=17$. $17 \div 5=3.4$. *$n<4$*

$n=3$: $21-3=18$. $18 \div 5=3.6$. *This is the number we have been looking for, so $3$ is the number that has been wiped out.*

Leo, Benny, Zach and Tyler from the University of Chicago Lab Schools had an alternative method for this question.

Step 1: We analyzed the set of numbers and knew that the crossed out number had to be lower than $4$. This is because the mean is higher than that of the original numbers, so the crossed out number has to be lower.

Step 2: We tried everything between $1$ and $3$ and found that $3$ was the correct answer.

Rubaiyat from Wilson's School was able to use a similar method to solve the next wipeout.

When a number has been wiped out, there are only $14$ numbers remaining, the mean of which is $7.\dot{7}1428\dot{5}$. Therefore the sum of the numbers left over is $14 \times 7.\dot{7}1428\dot{5} = 108$.

He then added up $1 + 2 + 3 + ... + 15$ using the formula that the sum $1 + 2 + 3 + ... + n$ of the first $n$ positive integers is $\frac{n(n+1)}{2}$.

The sum of numbers from $1$ to $15$ is $7.5 \times 16 = 120$. $120-108 = 12$.

Therefore the number crossed out was $12$.

For the last two wipeouts, Tess, Tilli and Clem, were able to use trial and error to obtain the solutions:

$N = 13$, and $9$ was wiped out.

$N = 51$, and $38$ was wiped out.

Hannah used divisibility to find the solution quickly. Her solutions are here.