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?
In this problem we were asked to take any number (less than
$1000$), add the squares of its digits, and then go on repeating
this until a pattern emerges.
You were asked to start at $145$, and you should have got the
sequence $$145 \mapsto42 \mapsto20 \mapsto4 \mapsto16 \mapsto 37
\mapsto58 \mapsto89 \mapsto145$$ and, of course, the sequence then
starts repeating itself.
After trying other starting points you should have been able to
guess that, whatever the starting point, eventually you will always
reach either $1$, when the number is called 'happy' (and then you
will stay at $1$), or the `cycle' given above, when the number is
You were also asked to show that whichever number we start with
(less than $1000$) we always get a number less than $1000$. Any
number less than $1000$ has only one, two or three digits, and each
of the digits is, of course, at most $9$. This means that when we
add the squares of the digits, the largest number we can get is $3
\times(9 \times9)$ which is $243$, and this
is less than $1000$.