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Calendar Capers

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?

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Reverse to Order

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?

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Card Trick 2

Can you explain how this card trick works?

Differs

Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2

Here are some examples from Robert Dunne and Tom Antnistle of St James Middle School, Bury St Edmunds.

(50, 6, 38, 6), (44, 32, 32, 44), (12, 0, 12, 0), (12, 12, 12, 12), (0, 0, 0, 0)

(86, 74, 34, 28), (12, 40, 6, 58), (28, 34, 52, 46), (6, 18, 6, 18), (12, 12, 12, 12), (0, 0, 0, 0)

With another lovely solution from Suzanne andNisha, (Y10) The Mount School, York.

We choose any 4 whole numbers, for example (100, 2, 37, 59) and take the difference between consecutive numbers, always subtracting the smaller from the larger, and ending with the difference between the first and the last numbers. When you repeat this process over and over again:

Eg: (100,2,37,59), (98,35,22,41), (63,13,19,57), (50,6,38,6), (44,32,32,44), (12,0,12,0), (0,0,0,0)

all sets of four numbers end up as four zeros. Sometimes it takes longer than others.

Eg: (58,8,15,40), (48,7,25,16), (6,4,10,8), (2,6,2,2), (4,4,0,0), (0,0,0,0)

Because the same term of the sequence is repeated over and over we say that the system ends in a fixed point (0,0,0,0).

Now let's try three numbers

(50,46,55), (4,9,5), (5,4,1), (1,3,4), (2,1,3), (1,2,1), (1,1,0), (0,1,1), (1.0.1), (1,1,0), (0,1,1), (1,0,1), ......

This pattern of two 1's and one zero continues for ever. The last 3 terms are repeated over and over again. They are called a 3-cycle.

(100,2,37), (98,35,63), (63,28,35), (35,7,28), (28,21,7), (7,7,21), (0,14,14), (14,0,14), (14,14,0), (0,14,14), ........

This time it's two 14?s and one zero and the pattern is slightly longer, but still goes on for ever ending in a repeating 3-cycle.

With five numbers:

(100,2,37,59,4), (98,35,22,55,96), (63,13,3,41,2), (50,20,8,39,61), (30,12,31,22,11), (18,19,9,11,19), (1,10,2,8,1), (9,8,6,7,0), (1,2,1,7,9), (1,1,6,2,8), (0,5,4,6,7), (5,1,2,1,7), (4,11,6,2), (3,0,5,4,2), (3,5,1,2,1), (2,4,1,1,2), (2,3,0,1,0), (1,3,1,1,2), (2,2,0,1,1), (0,2,1,0,1), (2,1,1,1,1), (1,0,0,0,1), (1,0,0,1,0), (1,0,1,1,1), (1,1,0,0,0), (0,1,0,0,1), (1,1, 0,1,1), (0,1,1,0,0), (1,0,1,0,0), (1,1,1,0,1), (0,0,1,1,0), (0,1,0,1,0), (1,1,1,1,0), (0,0,0,1,1), (0,0,1,0,1), (0,1,1,1,1), (1,0,0,0,1), (1,0,0,1,0), ......

Again this ends in a repeating cycle of period 15. Note that (1,0,0,0,1) is repeated so the cycle of 15 terms starting with (1,0,0,0,1) is repeated over and over again.

Some are quick: (5,12,26,7,1), (7,14,19,6,4), (7,5,5,5,3), (2,0,0,2,2), (2,0,2,0,0), (2,2,2,0,2), (0,0,2,2,0), (0,2,0,2,0), .... repeating 2's and zeros.

Nisha and Suzanne say "The only ones we could find with all zeros was with four numbers. There are all sorts of patterns in these others, but it's not easy to generalise them.''

Try 8 numbers and see what happens. Is it true that with 3 numbers you always end with a 3-cycle, with 5 numbers you always end with a 15-cycle and with 6 numbers you always end with a 6-cycle?