Wipeout
Take the numbers 1, 2, 3, 4, 5, 6 and choose one to wipe out.
For example, you might wipe out 5, leaving you with 1, 2, 3, 4, 6
The mean of what is left is 3.2
I wonder whether I can wipe out one number from 1 to 6, and leave behind five numbers whose average is a whole number...
How about starting with other sets of numbers from 1 to N, where N is even, wiping out just one number, and finding the mean?
Which numbers can be wiped out, so that the mean of what is left is a whole number?
Can you explain why?
What happens when N is odd?
Here are some puzzling wipeouts you might like to try:
The mean of what is left is 3.6
Which number was crossed out?
One of the numbers from 1 to 15 is wiped out.
The mean of what is left is $7.\dot{7}1428\dot{5}$
Which number was crossed out?
One of the numbers from 1 to N, where N is an unknown number, is wiped out.
The mean of what is left is $6.8\dot{3}$
What is N, and which number was crossed out?
One of the numbers from 1 to N is wiped out.
The mean of what is left is 25.76
What is N, and which number was crossed out?
With thanks to Don Steward, whose ideas formed the basis of this problem.
If you know what the average is, and how many numbers you have, what can you say about the total?
I wonder whether I can wipe out one number from 1 to 6, and leave behind five numbers whose average is a whole number...
Shaunak from Ganit Manthan, VicharVatika in India and Ahana, Ananthajith, Dhruv, Sehar, Saanvi, Nikhil, Insiya, Inaaya, Ishaan, Ishanvi, Abhay, Karthik, Vishnuvardhan and Rudra from Ganit Kreeda, Vicharvatika in India said it is possible. This is Shaunak’s work:
If I wipe out 1, then the mean of 2 + 3 + 4 + 5 + 6 = 20
20 / 5 = 4. 4 is a whole number.
If I wipe out 6, then the mean of 1 + 2 + 3 + 4 + 5 = 15
15 / 5 = 3. 3 is a whole number.
Luke from City of London School in the UK wrote an equation to find which numbers could be wiped out:
Sehar from Ganit Kreeda explained why using number bonds:
1, 6 are the only possibilities because there are three pairs of equal numbers which are 1 and 6, 2 and 5, 3 and 4. They all equal to 7, they are 3 pairs [which add up to] 7. You can have 6 or 1 because there are 2 remainders between 5 (the number of numbers we have to find the mean of). Ie 21$\div$5 = 4 remainder 1 or 3 remainder 6, so wiping out 1 or 6 leaves a multiple of 5.
Kavin from Hampton School in England used a formula for the sum 1+2+3+4+5+6 to find the numbers that can be wiped out (small correction in green):
The students from Ganit Kreeda explored what happens with other sets of six consecutive numbers. Click to see their work.
I tried this with:
2,3,4,5,6,7- Wipeout 2, mean is 5 (ALSO) Wipeout 7, mean is 4
3,4,5,6,7,8- Wipeout 3, mean is 6 (ALSO) Wipeout 8, mean is 5
I have deduced from my observations that when there is a set of 6 consecutive numbers, you can wipeout the first and last number to get the mean as a whole number.
Anathajith explained the 1st conjecture using a variable ‘a’.
He called the first number as a.
Then other numbers will be a+1, a+2, a+3, a+4, a+5.
So, the sum= 6a + 15.
If we subtract a or a+5 from this sum we will get 5a +15 or 5a +10 and both are multiples of 5
How about starting with other sets of numbers from 1 to N?
Aayan from Robert Gordon’s College in Scotland, Tireni from Eastcourt in the UK, 7a3 with Miss Kelly from Acklam Grange in the UK, Mugdha from Wallington High School for Girls and Ahana from Ganit Kreeda tried this out for different values of N. Aayan wrote:
1 2 3 4
I wipeout number one
2 3 4
To find the mean, I add 2+3+4 which is 9 and divide by how many numbers are left so 9$\div$3=3
From above I can also wipe 4
1 2 3
To find the mean I do 1+2+3=6 and divide it by 3 to get 2 as the mean.
Tireni also tried N=16:
Tireni described a rule about which numbers could be wiped out:
I noticed that whatever even number I chose, the first number(in this case one) and the last number ( in this case n), when you wipe that number out. The mean of all the other numbers is ALWAYS a whole number. Additionally, for example, if n=6, the mean if 1 was wiped out is 4. The mean if 6 was wiped out is 3.
Mugdha thought about why the rule works. Mugdha used a method that is very similar to Sehar's numbe bonds method (above). First, Mugdha noticed that, if you wipe out 1 or N when N is even, then you are left with an odd number of consecutive numbers. Then, Mugdha wrote:
In a consecutive number sequence, the mean is always the median (only if N (the last number) is odd).
This is actually also true if N is even, but in that case, the median/mean is not a whole number.
Mugdha showed some examples:
Mugdha used a diagram, similar to Sehar's number bonds method, to show why this happens:
Mugdha also used algebra to explain why, for even values of N, you can only wipe out the first or last number - not the others:
Wenhao from Hymers College in England, Luke, Kavin and Shaunak used a formula for 1+2+3+...+N to get the same result (you should wipe out 1 or N). Click to see Luke's work.
What happens when N is odd?
Aayan found numbers to wipe out when N is 5 or 7:
1 2 3 4 5
I wipe out 3
1 2 4 5
To find the mean, I add 1+2+4+5 which is 12 and divide by 4 which is 3.
1 2 3 4 5 6 7
I wipe out 4
1 2 3 5 6 7
To find the mean, I add 1+2+3+5+6+7 which is 24 and divide by 6 which is 4.
Sehar described a strategy:
When N is odd you have to wipe out the middle number ($a$) because $a$ is in the [middle.] There are some numbers more than it and some are less but they are equal[ly] more and less than it. They get cancelled. For example:
The students from Ganit Kreeda explained the rule in more detail:
Wenhao from Hymers College in England, Luke, Kavin and Shaunak used their formula for 1+2+3+...+N. This is Luke's work:
The students at Ganit Kreeda thought about other sets of consecutive numbers. Click to see their work.
Assuming N is odd and we need to make sequences from 1 to N, I have tried the following series of consecutive numbers.
I tried to make a sequence with 3 consecutive numbers first
- 1,2,3. When I wipeout 2, I get the mean as 2. I tried other combinations:
- 2,3,4- Wipeout 3, Mean is 6
- 3,4,5- Wipeout 4, Mean is 4
(Reason- Firstly, we know that when we wipeout one number there are two left, so the sum after wiping out should be a multiple of two. If we take the first number as variable $a$ then the rest of the numbers will be $a+1, a+2.$ So, the sum will be $3a+3.$ If we subtract the middle number ($a+1$) from it, we will get $2a+2,$ which is a multiple of 2)
Next, I tried to make a sequence of 5 consecutive numbers: 1,2,3,4,5 and found out that when we wipeout out 3, we get the mean as 3.
I tried these sequences too:
- 2,3,4,5,6- Wipeout 4, mean is 4
- 3,4,5,6,7- Wipeout 5, mean is 5.
Similar to 3 consecutive numbers I have deduced from my observations that when there is a set of 5 consecutive numbers, you can wipeout the middle number to get the mean as a whole number.
(Reason- Firstly, we know that when we wipeout one number there are 4 left, so the sum after wiping out should be a multiple of 4. If we take the first number as variable $a$ then the rest of the numbers will be $a+1, a+2, a+3,a+4.$ So, the sum will be $5a+10.$ If we subtract the middle number ($a+2$) from it, we will get $4a+8,$ which is a multiple of 4).
One of the numbers from 1, 2, 3, 4, 5, 6 is wiped out.
The mean of what is left is 3.6
Which number was crossed out?
Luke, Wenhao and Kavin set up a simple equation to find the answer. The equation uses the formula that they were using for the sum of consecutive numbers. This is Wenhao's work:
One of the numbers from 1 to 15 is wiped out.
The mean of what is left is $7.\dot{7}1428\dot{5}$
Which number was crossed out?
Again, Luke, Wenhao and Kavin set up an equation to find the answer. This is Wenhao's work:
One of the numbers from 1 to N, where N is an unknown number, is wiped out.
The mean of what is left is $6.8\dot{3}$
What is N, and which number was crossed out?
Kavin looked at the numbers and worked it out by considering the possibilities:
Wenhao and Shaunak both worked on similar methods to solve this and the final puzzle. This is Wenhao's method, and its application to the puzzle above:
Turn the given means into a mixed number or decimal, then write the whole numbers that encompass it, e.g. 3.7 would be 3 $\rightarrow$ 4. Once this is found, double it (since N$\div$2 you need to double to find N) so 3$\rightarrow$4 would become 6$\rightarrow$8. Using the denominator of our general formula, we can see that to find N we need to add one (opposite of N$-$1) so that it can multiply
into a whole number when multiplied with a fraction, (check step 4 and 5 of q1) thus finding N, then you can substitute it back into the formula and get your answer.
Below is the final puzzle, with Shaunak's method and solution:
One of the numbers from 1 to N is wiped out.
The mean of what is left is 25.76
What is N, and which number was crossed out?
To find the number that is wiped out, we have to know N. N is either given or it can be found out using the procedure given below :
1) Round down to the nearest ____.0 or ____.5. These are two possibilities.
Use the following procedure on both the cases to find the number that is wiped out:
1) Multiply the mean by N – 1. This is the sum of the remaining numbers. Call this number S.
2) Subtract S from N(N + 1)/2. The difference is the number that is wiped out. If the mean is a non-repeating decimal or a whole number, but the answer is a non-terminating decimal, then the case taken is wrong. Repeat the same process for the other case.
Note: If the mean is non-terminating, repeating decimal, then make the repeating part repeat 2 or 3 times, and round off the difference.
Ans: N is 51, and 38 is crossed out.
Can you follow Shaunak's method to get to Shaunak's answer?
Why do this problem?
By working on this problem, students will develop their understanding of the mean, and will be prompted to reason, generalise and justify.
Possible approach
"Take the numbers 1, 2, 3, 4, 5, 6 and choose one to wipe out. Then find the mean of your remaining numbers."
Collect students' answers.
"Oh, that's interesting, two people (one person/no-one) got a whole number answer!"
"We're going to explore the possible answers for the mean when you wipe out one number, from the set of numbers from 1 to N, where N is even.
Is it always possible to wipe out a number and get a whole number answer for the mean?
Can it be done in more than one way?"
You may wish to give different groups of students different values of N to explore, or invite them to choose their own (even) values of N. Once they have explored, bring the class together to share findings.
There are several ways to prove that rubbing out 1 or N gives a whole number mean when N is even. Once either end is wiped out, we are left with an odd number of consecutive numbers, whose mean will be the middle number. This can be proved algebraically, or by considering an image such as the one below:
This picture, which can be generalised, shows that for a set of five consecutive numbers, the mean is equal to the median.
When you remove the largest value, N, the remaining mean is $\frac{N}{2}$.
When you remove the smallest value, 1, the remaining mean is $\frac{N}{2}+1$.
Since all the other possible means will lie between the two consecutive means $\frac{N}{2}$ and $\frac{N}{2}+1$, there can be no other whole number solutions.
Students can also explore the case where N is odd; in this case there is only one whole number solution, as all the means lie between $\frac{N}{2}$ and $\frac{N}{2}+1$.
Finally there are some puzzling questions (of increasing difficulty) that students can have a go at.
Key questions
For the "puzzling wipeouts":
If you know what the average is, and how many numbers you have, what can you say about the total?
If you know the average, what are the possibilities for how many numbers you started with?
If the average has a recurring decimal, how can this help you to work out what number you divided by?
Possible support
M, M and M and Unequal Averages are suitable problems for students to try in preparation for this problem.
Possible extension
Invite students to consider what the mean average could be when wiping out numbers from different sets such as:
- $71, 72, 73, 74, 75, 76$,
- $2, 4, 6, 8, 10$,
- $1000, 1004, 1008, 1012, 1016$