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!