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?
Make a set of numbers that use all the digits from 1 to 9, once and
once only. Add them up. The result is divisible by 9. Add each of
the digits in the new number. What is their sum? Now try some other
possibilities for yourself!
What happens to the perimeter of triangle ABC as the two smaller
circles change size and roll around inside the bigger circle?
It was good to see so many attempts at the solution to this
problem. Try to remember that it is important to explain to the
reader each of the steps you take. Whilst many of you used some
element of trial and improvement in your method, which is fine, you
did not explain some of the things that you worked out before you
started trying different values. For example, many of you knew that
E = 1 and it was worth explaining why this was the case. In the
same way you were able to state that B=9 but only a few of you
tried to tell me how you knew this. Look at the detail in the
solution below, it may help you next time.
Finally, if several of you are sending in solutions from a
single school it might be worth looking at what each of you has
done to see if there are any patterns that you could use to extend
your answers and then send in a joint solution. Often the result
will be greater than each of the parts and it is good to talk about
what you are doing.
Clement Goh of the River Valley School, Singapore;
Elisabeth Elvidge, Alice Lewin and Alice Unwin of the Mount School,
Danielle Yule, Laura Lancaster, Briony Pollard and Emma Dellany of
the Mount School, York;
Prateek Mehrotra of Riccarton High School, New Zealand;
Andrei Lazanu of School number 205, Bucharest;
for identifying all the possible values for the digits. I have used
Andrei's, Prateek's and Clement's solutions as the basis of what is
Solution We know that E =1 because this is the
greatest amount that can be carried in an addition sum with two
From the units:
From the hundreds:
1. If D = 0
But B ?F, so D can't be 0.
2. D?1 because E = 1
3. If D = 2
F can't be 1, because E ?F so D ?2
4. If D = 3
Trying all possible combinations for A + C = 12:
5. If D = 4
Trying all possible combinations for A and C:
6. If D = 5
=> A + C =14
7. If D = 6
=> A + C = 15
8. If D = 7
=> A + C = 16
There are 12 solutions. These are all the solutions of the
problem. They are written in the following table: