You may also like

problem icon

Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

problem icon

Have You Got It?

Can you explain the strategy for winning this game with any target?

problem icon

Pair Sums

Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?

Split Clock Face

Age 11 to 14 Short Challenge Level:

If the sums of the numbers in each of the parts are equal, then the sum of the numbers in each part must be $\frac{1}{3}$ of the sum of all of the numbers on the clock face.

The sum of all of the numbers on the clock face is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78, and 78$\div$3 = 26. So the sum of the numbers in each part must be 26.

2 of the 3 parts will consist only of numbers that are next to each other on the clock face, as shown in the example below, so we need to find touching numbers on the clock face that add up to 26.



Finding numbers that work
Either the 1 and the 12 will be in the same part, or the 1 and the 12 will be at the ends of their parts. This would mean that there would be a section of touching numbers starting either 1, 2, 3, ... or 12, 11, 10, ... that add up to 26.

Trying to make a section of touching numbers starting 12, 11, 10, ... which add up to 26, 12 + 11 = 23 not 26, and 12 + 11 + 10 = 33 not 26. So that is not possible.

Trying to make a section of touching numbers starting 1, 2, 3, ... which add up to 26, 1 + 2 + 3 + 4 + 5 + 6 = 21 not 26, and 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 not 26. So that is not possible either.

So the 12 and the 1 must go in the same part. 1 + 12 = 13, so they need to be in a part with other numbers that also add up to 13. 11 + 2 is a possibility, and so is 6 + 7.

Using 6 + 7 would give the picture below, from which we can see that the other parts don't add up to 26. 


So 11, 12, 1 and 2 must form one of the parts, as shown.
  

3 + 4 = 7 and 10 + 9 = 19, so 3 + 4 + 9 + 10 = 26, so 3, 4, 9, 10 could make up another part. In that case, 5, 6, 7, 8 would make up the third part. 5 + 6 + 7 + 8 = 26, so the clock face can be split up as shown below.




Using algebra
Of the two sections of touching numbers on the clock face that add up to 26, at most one of them will contain the 12 and the 1. So at least one of them will not contain the 12 and the 1, so at least one of them will be a set of ascending numbers (that go up by 1 each time).

If the smallest of the numbers is $n$, then the next one will be $n+1$, the next one will be $n+2$ and so on. So we need either $n+(n+1)=26$, or $n+(n+1)+(n+2)=26$, or $n+(n+1)+(n+2)+(n+3)=26$, and so on.

If $n+(n+1)=26$, then $2n+1=26$, so $2n=25$ so $n=12.5$, which is not a number on the clock face.

If $n+(n+1)+(n+2)=26$, then $3n+3=26$, so $3n=23$ so $n=7.\dot{6}$, which is not a number on the clock face.

If $n+(n+1)+(n+2)+(n+3)=26$, then $4n+6=26$, so $4n=20$ so $n=5$. So $5 + 6 + 7 + 8 = 26$, and so one of the lines could go here.

Now choosing where the other line might go, note that $9+10=19<26$ but $9+10+11=30>26$, so 9 and 10 must go with some numbers from the right hand side of the clock face. $9+10+4=23<26$ but $9+10+4+3=26$.

That would leave $11+12+1+2=26$ as well. So the clock face can be split up as shown below.



Note: this method gives us a way to check that there is no other way to do this.

If $n+(n+1)+(n+2)+(n+3)=26$, then $n=5$, and we have one way to split up the clock face.

If $n+(n+1)+(n+2)+(n+3)+(n+4)=26$, then $5n+10=26$, so $5n=16$ so $n=3.2$, which is not a number on the clock face.

If $n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=26$, then $6n+15=26$, so $6n=11$ so $n=1.8\dot{3}$, which is not a number on the clock face.

If $n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)+(n+6)=26$, then $7n+21=26$, so $7n=5$ so $n=\frac{5}{7}<1$. There are no numbers less than 1 on the clock face, so, since adding more numbers will only make $n$ smaller, this will not work for any other sums of consective numbers.

You can find more short problems, arranged by curriculum topic, in our short problems collection.