Choose two digits and arrange them to make two double-digit
numbers. Now add your double-digit numbers. Now add your single
digit numbers. Divide your double-digit answer by your single-digit
answer. Try lots of examples. What happens? Can you explain it?
What are the missing numbers in the pyramids?
A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you the last two digits of her answer. Now you can really amaze her by giving the whole answer and the three consecutive numbers used at the start.
Mr Skipper (who didn't send us his first name)
was the first to send in a solution:
Call this $n$.
n n+1 n+2
n+1 n+2 n+3
So the product of the top left and bottom right is
The product of the top right and bottom left is
(n^2 + 4n + 4) - (n^2 + 4n) = 4.
Annette aged 13 did really well and
generalised the problem still further. She considered what would
happen if you had any sized square on the same addition grid.
Annette talks about the size of the square to
be N by N, and the first number in the top left of the square to be
since k=number in top left then