Choose any three by three square of dates on a calendar page.
Circle any number on the top row, put a line through the other
numbers that are in the same row and column as your circled number.
Repeat this for a number of your choice from the second row. You
should now have just one number left on the bottom row, circle it.
Find the total for the three numbers circled. Compare this total
with the number in the centre of the square. What do you find? Can
you explain why this happens?
Take any two digit number, for example 58. What do you have to do to reverse the order of the digits? Can you find a rule for reversing the order of digits for any two digit number?
Can you explain how this card trick works?
This problem looked at permutations or combinations of
possibilities. At first it might seem, as several people thought,
the eight runners could each finish the race in just one of three
positions -- which they wrote down as 8x3. This gave 24 possible
ways of finishing. In fact it is more complex problem than
The following solutions show that in fact any one of the eight
runners could come first, leaving seven runners who could take the
second place and any one of the remaining six could come in
The members of Burgoyne Maths Club explain
their different approaches to organising the information and how
they began to find a solution by writing down the possible ways of
the runners might finishing the race.
"Some of us gave the 8 pupils names, some letters and some
numbers. We started with listing the combinations with runner
number 1 finishing in first position, and showing where the other
runners could have finished".
Phew!! Writing out solutions really helps you see if all of the
possibilities have been included.
Then the Burgoyne Maths Club began to work on
the possibilities if runner number 2 was the one who came in at the
first position: 213
"At this point we realised there were 42 positions for each set
of results". Once they had discovered a pattern, the Club members
were able to finish the problem:
The total possible ways of coming first, second and third must
Is this method different? Bryan Hooi , of
Primary 5J at Henry Park Primary School described his solution way
of finding the solution.
"There are 8 possibilities for the first place.
For each of these 8, there are the other 7 possibilities for second
Therefore, for first and second place, there are 8 x 7 = 56
For each one of these, there are 6 possibilities for third
Therefore, for first, second and third place, there are 56 x 6 =
George Vassilev from Rosebank Primary School in
Leeds had another method. Can you see how is it similar or
different from the first two ways?
"First I named all the people = 1, 2, 3, 4, 5, 6, 7, 8.Then I
found out how many possible ways can they come in first, second,
and third if the first person is 1. So if the first person is 1
then there are seven people who can be second. So that leaves only
six people to be third. Now I multiplied six, seven, and one
6 * 7 * 1 = 42
That means there are 42 different ways for the people to come in
first, second, and third if the first person is 1. Eight different
people can be first and I have written about one of them but they
all have 42 different ways so that means I times 42 by 8.
42 * 8 = 336
So there are 336 possible ways of the people coming in first,
second, and third places".
Matthew Tattler and Steven
Townend from Moorfield Junior School explained their
solution differently and in a short way:
"There are eight possible people or ways of coming first. There
are seven ways of coming second, and there are six ways of coming
third. So that makes the sum 8x7x6, which equals 336, the total
number of possible ways".
The same answers were reached by different methods. I wonder if
anybody can add an explanation to Matthew and Steven's work that
shows how they arrived at their answer.