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 is odd).
B = A + 20 = the product of a prime number and a square number
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 square)
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 solution:
"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 http://www.nrich.maths.org/askedNRICH/edited/4087.html