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Clock Face

Age 7 to 11 Challenge Level:


1) Can you draw a straight line across the centre of a clock face so that the numbers on both sides of the line have the same total? Can you do this another way?


Andy, Courtney, Lukas and others solved this by looking at the sum of the numbers on the clock face, and then trial and error. Lauren writes;

By adding up all the numbers on the clock face, I got the total $78$. I then divided $78$ by $2$ as we are dividing the clock face into two equal parts. This equals $39$. Then by adding up six consecutive numbers around the clock face I found that the line should go between $9$ and $10$, and $3$ and $4$. As $4+5+6+7+8+9=10+11+12+1+2+3=39$.

A group from Archers Court in Dover solved the problem using an algebraic method. This method also helps the second question, on whether there is another solution.

The sum of all the numbers on the clock face is $78$, so you want to find six consecutive numbers that add up to $78\div2=39$. If we call the first number $n$, then the sum of six consecutive numbers is $n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=39$. By now gathering terms we get the equation $6n+15=39$ So $6n=24$ and $n=4$ Then the six consecutive numbers are $4,5,6,7,8,9$ and so the line runs from after $9$ to after $3$. As there was only one solution to the algebraic equation, then this is the only solution to the problem.

2) Can you draw two lines (like the hands of the clock) to divide the clock face so that the total of the numbers on one side of the lines is twice the total on the other side? Can you do this in another way?

Junnrui Hu wrote;

I realized that the total should be divided into three parts, so the two of the parts together would then be twice as large as the other part. $78\div3=26$. I then had to find consecutive numbers that added to $26$. By choosing a number and adding up around the clock from it until I reached $26$ or went over and missed $26$, I found that $11+12+1+2=5+6+7+8=26$. So the line can go from between $10$ and $11$ to between $2$ and $3$ or from between $4$ and $5$ to between $8$ and $9$.

3) Can you divide the clock face so that that the total on one side of the lines is five times more than the total on the other side? Can you do this in another way?

Harry from the Beacon School solved this using an algebraic method.

The sum of all the numbers is $78$. If we call $x$ the total on the small side, then $5x+x=78$, so $6x=78$ and $x=13$. So the sum on the small side is $13$. Now we must find consecutive numbers that add up to $13$. These are only $12+1$ and $6+7$. So there are two places for the line, from after $11$ to after $1$ and from after $5$ to after $7$.

4) Can you draw two lines to divide the numbers so that the total of the numbers on each side of the lines are both multiples of six? In how many different ways can you do this?

Lily and Amy from Bow Brickhill School solved this problem using a method of trial and error, by adding consecutive numbers until they totaled a multiple of $6$. They found that $9+10+11+12=42 (7\times 6), 1+2+3=6 (1\times 6), 4+5+6+7+8=30 (5\times 6)$. Here they have found three groups that sum to a multiple of six, so any combination of these will work, say having the lines after $8$ and after $3$, or after $8$ and after $12$, or after $12$ and after $3$.

Susanna from St George's also used a trial and error method to find several solutions.

Junnrui Hu wrote;

To make both sums a multiple of six, I tried to find consecutive sequences which added up to multiples of six. First I used sequences of length $1$, which is just $12$ and $6$ on their own. These can be isolated by the lines, and as $78$ is a multiple of $6$, if you take away by isolating a multiple of $6$, the rest will also be a multiple of $6$. Then looking at sequences of length $2$ there are none as the sum is always odd. Trying three adjacent terms there is a pattern $1+2+3=6=2\times 3=6\times 1, 2+3+4=9=3\times 3, 3+4+5=12=4\times 3=6\times 2$. So the groups $(1,2,3), (3,4,5), (5,6,7), (7,8,9), (9,10,11)$ and $(11,12,1)$ all are multiples of $6$.\
With 4 adjacent terms I found a pattern that starting on every third number will work, so the groups $(3,4,5,6), (6,7,8,9), (9,10,11,12)$ and $(12,1,2,3)$. In the same method I tried 5 consecutive numbers and found $(4,5,6,7,8)$ and $(10,11,12,1,2)$. There are none with $6$ adjacent numbers as this sum is an odd number as there are $3$ odd numbers in $6$ adjacent numbers.
And so you can find the solutions by isolating the groups, $(6), (12), (1,2,3), (3,4,5),$ $(5,6,7), (7,8,9), (9,10,11),$ $(11,12,1), (3,4,5,6),$ $(6,7,8,9), (9,10,11,12),$ $(12,1,2,3), (4,5,6,7,8)$ and $(10,11,12,1,2)$. So there are $14$ ways of making to groups that add to multiples of $6$.

5) Can you draw two lines so that the numbers on each side add to a prime number? Can you do this in another way?

Junnrui Hu answers this;

I had to think of two odd prime numbers that added up to $78$ (since $2$ is the only even prime). First I wrote all the primes from $1$ to $78$. $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73$. I found 7 pairs of primes that add to $78$, $(5, 73), (11, 67), (17, 61), (19, 59), (31, 47), (37, 41)$. Now I look at ways of making the smaller number with consecutive numbers. There are two ways of making $5 (5, 2+3)$, two way for $7 (3+4, 7)$, two for $11 (11, 5+6)$, one for $17 (8+9)$, one for $19 (9+10)$, and none for $31$ and $37$.
This gives $8$ different ways of diving up the clockface to give two sides where the sums are prime numbers.

Well done to Junnrui Hu for his excellent answers, attempting every single question. Congratulations also to Ian who solved every question using just his knowledge of triangular and prime numbers.

The last question remains open to anyone who wishes to investigate it, can you find any other interesting ways to group the numbers on a clock face by drawing two lines?