### Bat Wings

Two students collected some data on the wingspan of bats, but each lost a measurement. Can you find the missing information?

Kyle and his teacher disagree about his test score - who is right?

### Kate's Date

When Kate ate a giant date, the average weight of the dates decreased. What was the weight of the date that Kate ate?

# Wipeout

##### Age 11 to 16 Challenge Level:

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 $\frac{N+1}{2}$.

 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.

Tinotenda from Bishop Ullathorne School and Sammi and Harvey from Wormholt Park School gave good solutions to the first puzzling wipeout. Here is Tinotenda's solution:

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.