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?
Take any whole number between 1 and 999, add the squares of the
digits to get a new number. Make some conjectures about what
happens in general.
You may have found that this trick does not work if the fourth
card from the bottom is the same numerical value and colour as one
of the 3 chosen cards or as the first, second or third card from
the bottom, a probability of 6/45. When the 'magician' looks at the
cards and sees that this has happened the best thing is to carry on
with the trick but first to say that the cards should be shuffled
again and give some convincing reason!
The card which the volunteer keeps will always be the fourth
card from the bottom of the pack which has the same numerical value
and colour as the predicting card. This is because, whatever 3
cards are selected by the volunteer, with these 3 cards and the
predicting card, 4 cards are removed from the pack. Then 45 cards
are counted out, and this leaves the last 3 cards to make up 52
altogether. Suppose the 3 cards selected have values x ,
y and z then the number of cards counted out is
(15 - x ) + (15 - y ) + (15 - z ) +
x + y + z = 45.
Correct solutions were sent in by:
Sarah - Archbishop Sancroft High School