Sam displays cans in 3 triangular stacks. With the same number he
could make one large triangular stack or stack them all in a square
based pyramid. How many cans are there how were they arranged?
Here is a collection of puzzles about Sam's shop sent in by club
members. Perhaps you can make up more puzzles, find formulas or
find general methods.
Investigate polygons with all the vertices on the lattice points of
a grid. For each polygon, work out the area A, the number B of
points on the boundary and the number of points (I) inside the
polygon. Can you find a formula connecting A, B and I?
We received correct solutions from Julia Collins, from Langley
Park School for Girls in Bromley, Ross Haines, Ben Young and Philip
Bennington from Mosgiel Intermediate School, and Andrei Lazanu from
School 205 in Bucharest, Romania. Well done to you all.
Julia Collins said:
"I think I've found a solution to this problem, although I'm
sure it's not the best solution, so I will be interested to see
what other people have come up with!
We know that A + 100 is an even number, so we know that A is
even. Thus we know that one of the primes making up each of the
numbers A, B, C and D is 2 (because any other combination of primes
B = A + 20 = the product of a prime number and a square
Since A is even, let the prime be 2. Then the only square number
that gives a sensible date is
31 2 = 961.
So B = 1922.
So I checked the other numbers to see if the solution was valid,
and it was!
A = 1902 = 2 x 3 x 317 (three primes)
B = 1922 = 2 x 961 = 2 x 31 2 (prime and a
C = 1982 = 2 x 991 (two primes)
D = 2002 = 2 x 7 x 11 x 13 (four primes and even)
I don't like the solution because it involves a guess, and so
doesn't show whether the solution is unique or not. Can it be done
purely by logical thinking?"
Andrei Lazanu was able to show that this was a unique
"First I observed that the data known for the problem (A, B, C,
D) corresponds to A, A + 20, A + 80 and A + 100.
Because A + 100 is even, all the other terms are also even
(because 20, 80 and 100 are even).
Now, I write the data I have:
A = 2 * prime * prime
B = A + 20 = 2 * x 2
C = A + 80 = 2 * prime
D = A + 100 = 2 * prime * prime * prime
I start by considering the condition for B and observe that x
must be smaller than 32 (since 2 * 32 * 32 = 2048):
I stop here because from now on A is below 0.
So, 1902 is the unique year that matches all conditions for Great
Granddad's birth ."
PS. Julia Collins wrote to us pointing out a slight flaw
in Andrei's argument. They discussed this on the web-board, and
their discussion can now be read at