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?
Another crop of good solutions! Robbie said he
and his Mum nearly lost their marbles over this one. Quite a number
of you reasoned that there are the more yellow marbles than any
other colour. All three solutions were found by Josh of Alameda
School Ampthill (well done Josh!), by Y9 of the Mount School, York
as a team effort, by James of Hethersett High School, who used a
spreadsheet to good effect and, from the same school, by Rachel,
Geoffrey, and Sarah and also by the Key Stage 3 Maths Club,
Strabane Grammar School, N. Ireland. The most carefully reasoned
argument covering all possible cases was done by Xin Ying from Tao
Nan School, Singapore.
Let R, G, Y and B be the number of Red, Green, Yellow and Blue
There are 4 conditions to fulfil: (1) R + G + Y + B = 12; (2) R
> G; (3) (G+B) > R;
(4) (Y+G)> (R+B). We shall use logical thinking to get the
From condition (3), (R+B) is at most (12¸ 2) - 1 =
So (R+B) could be 5, 4, 3, 2, or 1.
Case 1 : (R+B)=1
Impossible as R and B are both not zeros.
Case 2 : R+B = 2
Then, R=1 and B=1. But R> G and G is not zero. So this case is
Case 3 : (R+B) = 3
Then (R+B) = 1+2 or 2 +1. But it must be that R=2 and B=1 so as
to fulfil condition (2). Now, R> G tells us that G=1. But (G+B)
is not greater than R. Condition (3) is violated. So this case is
Case 4 : (R+B) = 4
(a) R+B = 1+3 or 3+1. It must be that R=3 and B=1 from condition
(2). Since R> G, G=2 or 1. Either case, (G+B) is not greater than
R. So this case is impossible.
(b) R+B=2+2. Then R=2 and B=2. Since R> G, G=1. Then, Y=7.
Upon checking, all 4 conditions are fulfilled. This case works!
Case 5 : (R+B)=5
(a) R=2 and B=3. Since R> G, G=1. So, Y=6. Upon checking, all
4 conditions are fulfilled. This case works!
(b) R=3 and B=2. Since R> G, G=2 or 1. If G=1, (G+B) is not
greater than R. This case is not possible. So G=2. Then Y=5. Upon
checking, all 4 conditions are fulfilled. This case works!
(c) R + B = 4 + 1 (1+4 is impossible). So R=4 and B=1. Since
R> G, G = 3, 2 or 1. But either case, (G+B) is not greater than
R. This case is impossible.
In short, only 3 situations are possible:
5 yellow marbles, 3 red marbles, 2 blue marbles and 2 green
7 yellow marbles, 2 red marbles, 2 blue marbles and 1 green
6 yellow, 2 red, 3 blue and 1 green marbles.