A Mixed-up Clock
Problem
A Mixed-up Clock printable sheet
Here is a clock face where the numbers have become all mixed up. In the picture, each of the numbers is represented by a letter. Can you find out which letter represents which number using the ten statements below?
- No even number is between two odd numbers.
- No consecutive numbers are next to each other.
- The numbers on the vertical axis (a and g) add to $13$.
- The numbers on the horizontal axis (d and j) also add to $13$.
- The first set of $6$ numbers (a, b, c, d, e, f) add to the same total as the second set of $6$ numbers (g, h, i, j, k, l).
- The number at position f is in the correct position on the clock face.
- The number at position d is double the number at position h.
- There is a difference of $6$ between the number at position g and the number before it (f).
- The number at position l is twice the top number (a), one third of the number at position d and half of the number at position e.
- The number at position d is $4$ times one of the numbers next to it.
Getting Started
Can you now find some more that follow from this one?
Student Solutions
This challenge brought in over 150 submissions. A large number of solutions came from different parts of the world, namely, Bejing, Honk Kong, Dubai, Australia, Great Britain and the United States of America.
So here are a few submissions that reflect the kind of thinking that went on:
From Ottilie, Lucy and Grace at Hordle Walhampton School Lymington:
A = 2, B = 9, C = 3, D = 12, E = 8, F = 5, G = 11, H = 6 , I = 10 , J = 1, K = 7, L = 4
What did you do first?
We saw that one number was definite on the clock face (letter F). We then found that the instructions said G related to F with a difference of 6. Which meant we could then find the vertical numbers.
How did you work through the problem?
We worked through all the statements and did not do them in order. Some statements were not possible to solve until the very end.
What were the most difficult bits?
Trying to see how the letters connected with the numbers
How did you check your answer was right?
We went through each statement to see if it was true and then checked our answers with two teachers.
From Sarah at Yew Chung International School Of Beijing:
The first step of this problem is to find any definite numbers, which I found on clue 6. (Number position F is in the correct position on the clock face.) So now we know that F is 5.
Next we drew a diagram to remember what we did, we recorded our information on it.
From clue 2 (No consecutive numbers are next to each other) I know that 1-12 a-b are not in order.
From clue number 1, it gives us a good clue (No even number is between two odd number) so I know that E or G are even numbers or odd numbers. One must be even and one must be odd.
I know that G would be number 11. This is because on clue 8, (There is a difference of 6 between the number at position G and the number preceding it (F)) G must be 11 because if 5 minuses 6, it would be negative, there is no negative number on the clock. Also, 5 plus 6 equals 11.
You can figure out A because you know G. On number 3 it says the numbers on the vertical axis A and G add up to thirteen. So if G is 11, all you have to do is minus 11 from 13, which is 2.
Question 9 is a good clue to do because you can find three numbers. (The number at position L is twice the top number A, one third of the number at position D and half of the number at position E.) First, you know what number A is so you can find L which can bring you to the letters D and E.Since A is 2, twice A is 4 which means L is 4.
To figure out D, it says that L is one third of D so you have to times L(4) by 3 which equals 12.
To get E, it says that L is half of D so you have to let L(4) times 2 which equals to 8.
Now you have the numbers of the letters L, D and E.Numbers H and I are definitely even numbers because so far, the pattern is even even odd odd and because on clue 1, it says No even number is between two odd numbers. To get the number for letter H, you need clue number 7. (The number at position D is double the number at position H.) Which also means D is half of H. If D is 12, then half
of twelve is 6, which means
that H is 6.
Number 4 helps us to get the number of the letter J. On clue 4, it says (The numbers on the horizontal axis D and J add up to 13.) If D is 12, 13 minus 12 equals 1 which means that J is 1.
On clue 10, It tells us that the number position D is 4 times one of the numbers adjacent (next) to it. C and E are next to D. Since we already have E, then we need C. If D is 12, then 12 divided by 4 equals to 3. C is 3.
There is only K, B and I left. We know that I is definitely an even number because the order is even even odd odd, this is because clue one says that no even number is between 2 odd numbers. The only numbers left are 7, 9 and 10. There is only one even number so I must be 10.
Lastly, clue 5 helps you with the total answer. Clue 5 tells us that the first set of 6 numbers A-F add to the same total as the second set of 6 numbers G-L. We don't have all the numbers yet but you can still add them together to get the other numbers. A-F missing 1 number, (the number we have) 2, 3, 12, 8, 5. If you add them together, it equals to 30. G-L is missing 1 number,(the numbers
we have) 4, 1, 10, 6. If you add them together, it equals 32. We need the numbers 7 and 9. 7 and 9 have the difference of 2. 32 and 30 also have the difference of 2. If A-F only has 30 and G-L has 32, then the greater number that is left(9) goes to B. The smaller number goes to G-L because without the number for letter K, G-L had a greater number after added up.
From Class 5 Peterchurch Primary School:
A 2, B 9, C 3, D 12, E 8, F 5, G 11, H 6, I 10, J 1, K 7, L 4
After reading through the clues carefully, we realised that clue six told us that as F was in the correct clock position, it must be number 5. Clue 8 told us that G was 11. Therefore we knew that A = 2 (as 11 + 2 make 13). Next, we worked out from clue 9 that L was twice A making this number 4, which led us to know that D was 12 (3 x 4) and L was half of E making it 8. Once
we knew D we could work out J, using clue 4 (this was 1). Clue 7 told us we could work out H which was 6. We worked out C from clue 10 (D is four times this number). Then we added all the numbers in the clock face together making 78 and divided this by 2, equalling 39. We split the clock as per the instructions and added the numbers together. A, C, D, E and F = 30 so we knew B had
to be 9. Finally we knew no consecutive numbers were next to each other so 7 couldn't go next to H, therefore K = 7. Leaving I as number 10. We solved the problem!
Finally from Adi at Hymers College Junior School:
Basically I am going to tell you my solution in steps.
Step 1 Well I first read every clue there was. Then I looked at clue 6 which said F is in the correct position on the clock so it had to be 5. Now I looked at clue 8 which said that G is 6 more than the one next to it so G must be F+6 and that would =11 because fFwas 5 and G was 6 more than F.
Step 2 After that clue 3 says that the vertical axis (G and A) adds up to a total of 13, so if G =11 that means that A is 13-11= 2 so now you have worked three of the twelve numbers.
Step 3 Now I looked at clue 9 which said L is twice the amount of A so A= 2, to find L I did 2x2=4 so now we have solved the letter L. After that it says that in clue 9 again that L is one third of D and to solve that you need to do 4x3=12 which is the number for D. Now in the same clue it tells you
that L is half of E so 4x2=8 would be the answer to D.
Step 4 Now the letters I still had to solve are K, B, C, J, I, H. First to find J it clearly says in clue 4 that the horizontal axis (J and D) that they add up to 13 so if D=12 that means that you just need to add 1 to make 13 so J=1.bNow moving to clue 7 it says that H is 2x less than D so you just need to do 12 divided by 2 which= 6 which is the number for H.
Step 5 Now I looked at the last clue and that said that D is 4x the number next to it so if we have solved E it had to be C and to work this out you needed to do 12 divided by 4=3 and that's the number for C. Now is the trial and error part I had because the three letters K, I and B had no clues to solve these numbers for the letters. So I looked at what numbers were
missing from a normal clock and the numbers were 9, 7 and 10 but before working out the numbers for those three letters I first looked at clue 5 and that clue said that the letters A-F = the same amount as G-L, so I did the sum without the missing numbers and for A to F the sum added up to 30 and G to L the answer was 21.
Step 6 Now since I knew that the second set of numbers added up to 21 and they had two missing numbers so for letter K I gave the number 10 and I gave 7 to I. After that the total for the second set was 39 so if the first set of numbers added up to 30 then that means that B= 9. So now since both of the sets of numbers add together to the same total I realised I had solved
the problem.
Well done all of you, this was a tremendous effort in mathematical thinking.
Many years on in February 2020 we had this solution sent in from Emerson at Village Community School in New York
My approach was to read through all of the questions and to start with the easiest ones in order to eliminate possible numbers prior to tackling the harder problems.
The first step I did was #6 because it is the only clue that gives you a definite number. This placed 5 in position F.(eliminated 5)
I then move to clue #8 which says that position g had a difference of 6 from position F, which I knew had a value of five. Because no clock has negative numbers, I knew position g was 11. (eliminated 5,11)
This allowed me to tackle clue #3, since I knew that position g was 11 then position a was 13-11, so position a got a value of 2. (eliminated 2, 5, 11)
This allowed me to tackle clue # 9. Since 2 times 2 equals 4, I knew position l was 4. This also allowed me to know that position D was 3 times position l which gave me 12 and position e was 2 times position l which gave me 8. (eliminated 2, 4, 5, 8, 11, 12)
This allowed me to tackle clue #10 since the only open position next to d was c then I knew 12 divided by 4 was 3. (eliminated 2, 3, 4, 5, 8, 11, 12)
Clue #9 also made it possible to figure out clue #4 getting a value of 1 for position j since 13-12=1 (eliminated 1, 2, 3, 4, 5, 8, 11, 12)
Knowing the value of d, allowed me to solve clue #7. 12 divided by 2 is 6 which goes in position h. (eliminated 1, 2, 3, 4, 5, 6, 8, 11, 12)
There was only one last even number to pair with 6 at position h (clue #1) so that meant that position i = 10 (eliminated 1, 2, 3, 4, 5, 6, 8, 10, 11, 12)
The final clue that gave the last two numbers was #5. By adding the numbers I had on the clock I found that g to l equals 32 and a to f equals 30. That meant that the bigger number (9) needed to go into the open position b and the smaller number (7) needed to go to open position k. (eliminated 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. And complete!)
Thank you for that very clear and precise explanation, well done!
Teachers' Resources
Why do this problem?
This problem is one that needs logical thinking. At first it might look impossible but if the clues are followed in a suitable order then it only requires simple arithmetic to solve.
Possible approach
You could begin with a few oral challenges to remind the class of number properties such as odd, even and consecutive. For example, could they suggest three consecutive odd numbers greater than 50 but less than 70? How many different sets of consecutive odd numbers are there between 50 and 70?
Share the problem with learners and describe briefly what it entails without reading all the information at first. Invite pupils to talk in pairs about how they would start the problem. You could read through the clues in turn, stopping after each to give children chance to discuss what exactly could be learned from it. Give them time to look through each clue in pairs and decide what can be
filled in immediately. You could draw everyone together to share ideas before giving more time to complete the challenge.
A plenary could give learners the opportunity to explain the steps they took to solve the problem. Making it clear that the ordering of the information constitutes a systematic approach will help them begin to understand the meaning of being systematic and therefore apply it to future problems.
Key questions
Can you find a statement that gives you a definite number that you know is right?
Can you now find some more that follow from this one?
Possible extension
Learners could use the clock-face given to work out some clues for themselves.
Possible support
Some children may need help reading the clues so this may influence your choice of pairs.
Sealed Solution
Problem
Sealed Solution printable sheet
Here is a set of ten cards, each showing one of the digits from 0 to 9:
The ten cards are divided up between five envelopes so that there are two cards in each envelope.
The sum of the two numbers inside it is written on each envelope:
What numbers could be inside the "8" envelope?
Thank you to Alan Parr who inspired this task.
Getting Started
What are the possible ways of making the numbers on the envelopes?
Which number has the fewest possible combinations? It might be worth starting from this envelope and looking at what could be in the others.
Student Solutions
We had lots of solutions sent in for this activity, so thank you to everybody who shared their ideas with us.
Anika from the National Academy for Learning in Bangalore, India sent in a table of possibilities for the numbers in each envelope, with this explanation:
I made five columns, one each for the envelopes. For each number, I wrote down the possible pairs that could be in the envelope. I found that the envelope 8 can contain the numbers (0,8),(5,3),(1,7),(6,2).
Well done for finding all these possibilities, Anika. In Anika's table we can see all the pairs of numbers that make the total on each envelope.
Leo from Kings' School in Dubai, UAE used the same idea as Anika to draw a table of all the different possibilities. He then worked out which of these possibilities would work with each other, as the same number can't be in more than one envelope:
I started off by creating a table of all of the possible number bonds to each of the targets on the envelopes.
I then worked systematically so that I tried every potential pathway across the table. Some of them proved to be impossible but I was able to find all 3 solutions. I have also included an example of a pathway that was impossible; it was where I started the process.
I know that there are only 3 possible solutions as I worked through every possibility in a systematic fashion.
Well done for working systematically through all of the different possibilities, Leo. Leo's solution shows that the "8" envelope can contain 0 and 8, 5 and 3, or 1 and 7, but it looks like the cards 6 and 2 can't be in the "8" envelope. I wonder why?
James and Oais from the UK found the same solutions as Leo, and they explained why 6 and 2 can't be in the "8" envelope:
We made a chart with all the possibilities for making the numbers on each envelope, then we tested the different ways it could go.
If you start by looking at how to make 14, there are two: 5/9 or 6/8. Then, if you try to make 13, you can see that there is only one way that works for each. When you choose 5/9 for 14, you have to choose 6/7 for 13. When you choose 6/8 to make 14, you have to choose 4/9 to make 13 because you've used up the other numbers.
The numbers you have to make 8 are 0/8, 1/7, 2/6 or 3/5 but you'll never be able to use 2/6 because you always have to use the 6 to make 13 or 14. So the answer is 0,8,1,7,3 or 5!
Good explanation! I wonder if it's more helpful to start with the smallest numbered envelopes like Leo did, or to start with the largest numbered envelopes like James and Oais did?
Thank you as well to these children who sent in excellent solutions: James from Hamstel Junior School in England; Mia and Hari from Kings' School in Dubai; Anna from Sahuaro Elementary in the USA; Lauren from Australia; K from Crossflatts in the UK; Saanvi from Newcastle-under-Lyme in the UK; Isla from Walton and Holymoorside Primary in the UK; and Ci Hui Minh Ngoc from Kong Hwa School in Singapore.
Teachers' Resources
Why do this problem?
Sealed Solution offers the chance for children to work in a systematic way and is a great context in which to encourage them to explain and justify their reasoning.
Possible approach
Begin by familiarising children with the context: Using digit cards 0 to 5, invite the class to watch as you put 0 and 1 in one envelope and write their total on the outside (or on a 'post-it' note stuck to the envelope). Put 3 and 5 in another envelope, again writing their total on the envelope. Explain that the other two cards will go in the last envelope. What will the total be? How do they know?
Try this again, this time putting 0 and 5 in one envelope and recording the total. But then put two cards in another envelope without showing them to the children. Write the total on the outside of the envelope. Repeat this for the third envelope. (For example you could have 1 and 3 in the first and 2 and 4 in the second.) What numbers are in the two envelopes? How do they know?
Try again, this time keeping 0 and 5 in the first envelope but suggest that you want to put the other cards in pairs into the envelopes, so that the totals on the other two are the same. What could you do? How do they know? At each stage, children can be working in pairs, perhaps using mini-whiteboards and digit cards to try out their ideas.
Key questions
Which envelope shall we try first? Why?
What could be in this envelope?
Are there any numbers which you know definitely aren't in this envelope? Why?
Are there any other solutions?
Possible support
Having digit cards available for children to use will free up their thinking and will make it easier to try out different ideas without worrying about crossing ideas out on paper.
Possible extension
Children could make up their own problem along these lines.
Alan Parr, the creator of this task, wrote to tell us:
'I've recently returned to this for the first time in ages, working with some Year 6s [10 and 11 year olds]. They found it so accessible and involving that we took it to places I'd never previously dreamt of.' You can read what they did in the first April 2015 post on Alan's blog, and he writes about the task again in
two March 2017 posts.
Beads and bags
Half Time
Buying a Balloon
Cubes Here and There
Trebling
Factor Lines
Sealed Solution
Magic Vs
All the Digits
Finding Fifteen
Button-Up Some More
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
This Pied Piper of Hamelin
Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
Reach 100
Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100.
5 on the clock
Nine-Pin Triangles
Inky Cube
Sitting round the party tables
A Mixed-up Clock
What Do you Need?
A Square of Numbers
Light the Lights Again
Each light in this interactivity turns on according to a rule. What happens when you enter different numbers? Can you find the smallest number that lights up all four lights?
Prison Cells
Two Primes Make One Square
Tea Cups
Counting Cards
Curious number
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
Make 37
Four bags contain a large number of 1s, 3s, 5s and 7s. Can you pick any ten numbers from the bags so that their total is 37?