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?
There were several solutions sent in to this problem, which was about substituting numbers for letters. Most people interpreted the problem in the same way, that different letters could not share the same value, in other words H and E had their own value and E could not be the same as H. Well done to each of the solution finders.
Christina Ivanova, from Marlborough Primary School explained her strategy for keeping track of the digits she used and to help her discover the value for each letter.
I saw there were 10 different digits which meant there would be 10 different digits. From then on I started guessing the numbers, crossing off the digits as I used them. My final solution was:
Christina's solution was shared by Ece Tugc and Simin Araz (Irmak Primary School, Istanbul, Turkey)
A second solution was found by Sophie and Annabelle (Annesley College, Adelaide, Australia), Sinan Ersanli (from the sixth grade of Irmak Primary School, Istanbul, Turkey) and Ece Demir (also in sixth grade, in the Irmak, Private Primary School). Sophie and Annabelle wrote it out in the following way:
H = 4
F = 2
L = 5
R = 6
O = 0
V = 3
N = 8
E = 1
This is the sum with the letters replaced with the digits
Simin Araz from Irmak Primary School in Istanbul, Turkey found a third solution to the problem:
And Rose from Claremont Primary School in Tunbridge Wells submitted this solution, which was different to all the other ones that we had received.
Oliver, Amiko, Scott, Ben, Billee and Aidan, from Curdworth Primary School in the West Midlands, worked out one more solution:
T = 4
H = 6
R = 5
E = 1
F = 8
O = 2
U = 9
L = 0
V = 3
N = 7
Well done to you all.
Are there any more solutions that could be found for this problem?
Are there any factors that limit the number of possible solutions?
Can anybody suggest a strategy for finding all the possibilities?