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Thank you to everyone who submitted a thoughtful solution to this problem, we received a large number. Along with many others, Sofia from Caxton College in Spain explained the conditions on the five numbers:
First you look at the mean. Imagine your mean is 4 and you have to give 5 numbers. So you multiply 4 by 5. Your product is 20. So your 5 numbers have to add up to 20 as when you divide them you will get the number 4.
If your median is 3 you have to make sure that the number 3 is in the middle of all the 5 numbers as they go in numerical order (_,_,3,_,_) .
Before the number 3 you have to have smaller number as it goes in numerical order e.g (2,3,3,_,_) and after the number three the numbers have to be
bigger (2,3,3,4,7). Now your median is 3 and your mean is 4.
If your mode is also 3 then all you have to do is assure that you use the number 3 more than other numbers. Like in my example I used the number 3 two times, and the rest, one time. So my mode is 3.
Using these ideas, Saanvi from the UK, Haoyang, Suhani and Zoe, Robert and Gabor from Garden International School in Malaysia, N from The Vine Inter-Church Primary School in the UK, Rafferrty, Ben and Javier from Twyford School, Winchester in the UK, Saanvi from Open Sky International in France, Connor from Shirley Boys High School in New Zealand, Niko from Mount Grace in the UK, Yuxin from St James Senior Girls School, Nikhil, Abhay and Ishaan from Ganit Kreeda, Vicharvatika in India and Anna from Caxton College all sent one or more sets of 5 numbers that meet the criteria. Here are some of their examples:
Ben: 2,3,3,3,9
Saanvi from Open Sky International:
1. Mean of given set of 5 whole number is 4 = 3, 3, 3, 5, 6
2. Median of the Same set 5 whole number is 3 = 3, 3, 3, 5, 6
3. Mode of Same set of 5 whole number is - 3 = 3, 3, 3, 5, 6
Connor:
1+3+3+5+8
3,3 mode
3, median
1+3+3+5+8=20/5=4
4, mean
Yuxin found these sets, as well as some of the ones above:
1,2,3,3,11
2,3,3,4,8
2,3,3,5,7
3,3,3,3,8
3,3,3,4,7
Nikhil found this sets, as well as some of the ones above:
Suhani and Zoe found these sets, as well as some of the ones above:
Lyla from CHS South expressed the numbers algebraically, and then found some possible sets. Here is Lyla's work (click on the image to open a larger version). Can you match Lyla's sets to the sets above?
Students from Farlingaye High School in the UK, Harry from St Michaels Preparatory School in Jersey, Arjun from Garden International School, Liam from Great Valley Middle School in the USA, Davie from Rugby School Thailand in Thailand and Iris and Lola from CHS South worked systematically by considering the possibilities for the smallest number. This is Harry's work:
I then organised the numbers so the number one would be in the first position, leaving the last two digits to add up to 13.
The same then happened when I started at 2 as the first number, but the last 2 digits added up to 12.
And when I tried with number 3, the end 2 digits added up to 11.
If I put 4 as the starting number, then 3 wouldn’t be used as the mode, for the mode goes in numerical order.
Solutions:
1 2 3 3 11
1 3 3 3 10
1 3 3 4 9
1 3 3 5 8
1 3 3 6 7
2 3 3 3 9
2 3 3 4 8
2 3 3 5 7
3 3 3 3 8
3 3 3 4 7
3 3 3 5 6
Nina W from Berlin Cosmopolitan School in Germany, Jon from The Old Priory School in the UK, Aniket from The ABC International School in Vietnam, Ci Hui from Queensland Academy of Science Mathematics and Technology in Australia, Anathajith from Ganit Kreeda also worked systematically, by considering the positions of the 3s. This is Ananjith's work:
Suppose that when you arrange the five numbers in ascending order you get the number: a , b , c , d , e.
Case 1: c=3, d=3
a , b , c , d , e
| | | | | Only Possibility for this Case.
1 2 3 3 11
Case 2: b=3, c=3
a , b , c , d , e
| | | | |
1 3 3 6 7
1 3 3 5 8 5 Possibilities
1 3 3 4 9
2 3 3 5 7
2 3 3 4 8
Case 3: b=3, c=3, d=3
a , b , c , d , e
| | | | |
1 3 3 3 10 2 Possibilities
2 3 3 3 9
Case 4: a=3, b=3, c=3
a , b , c , d , e
| | | | |
3 3 3 5 6 2 Possibilities
3 3 3 4 7
Case 4: a=3, b=3, c=3, d=3 Only Possibility for this Case.
3 3 3 3 8
Note: The case where b, c, d and e are 3 is not possible because ‘a’ would have to be less than 3 and if it was, the sum wouldn't be 20.
Therefore, Total Number of possibilities = 1 + 5 + 2 + 2+1 = 11
Naiana and Heidi from Rubgy School Thailand, Natalie from Bangkok Patana School in Thailand, Ira from Sahyadri School KFI in India, Mohammed from Repton School Abu Dhabi in the UAE, Shaunak from Ganit Manthan, Vicharvatika in India, Ahana from Ganit Kreeda, Avni and Sihun from Garden International School, Ahana, Ananthajith, Dhruv, Sehar, Saanvi, Nikhil, Insiya, Inaaya, Ishaan, Ishanvi, Sanat, Abhay, Karthik, Vishnuvardhan, Rudra from Ganit Kreeda, listed all the possibilities systematically based on the total number of 3s. This is Avni and Sihun's work:
This is the Ganit Kreeda students' woek from the entire group. They included 0 as a positive number, so they have more solutions:
Umar from Eastcourt Independent School in the UK wrote some Python code to find the possible sets. Here is Umar's code:
import random # Using a set to store unique tuples unique_triples = set() for i in range(1, 100000): num1 = random.randrange(1, 13) num2 = random.randrange(1, 14 - num1) num3 = 14 - num1 - num2 # Create a sorted tuple to avoid permutations of the same numbers triple = tuple(sorted([num1, num2, num3])) # Ensure all numbers are unique and the tuple is not already in the set if len(set(triple)) == 3: unique_triples.add(triple) # Convert the set to a list if needed unique_triples_list = list(unique_triples) print(len(unique_triples_list)) print(unique_triples_list)
Can you work out what the code would output? Click to show an explanation.
Click to see Umar's investigation and conclusions based on the result of running the code:
James from Norwich School in the UK posed and answered another question:
I found a way to make a set of 5 numbers with the same mean mode median and range as long as it is even.