Man Food

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?

Sam Again

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.

Pick

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?

Letter Land

Stage: 3 Challenge Level:

We have had solutions to this problem from Alan, Atharv from Bedford Modern School, Andreifrom School 205 in Bucharest, Romania, Julia from Langley Park School for Girls, Clement from River Valley High School in Singapore, Christopher from St. Bees School in Cumbria, Rosamund from Riccarton High School in New Zealand and from Aditya. Well done to you all.

Everyone approached the question in a similar way. Aditya's approach follows:

We can simplify most of these equations first, before we start giving values to all of the letters.

As $A + C = A$ we know that $C = 0$.

As $F \times D = F$ we know that $D = 1$.

$B - G = G$, therefore $B = 2G$ (add $G$ to both sides of the equation).
$B / H = G$ therefore $B = HG$ (multiply both sides by $H$).

Since $B = HG$, and $B$ also equals $2G$,
$H$ must be $2$.

Since $B$ is twice $G$, the options are:

 $B$ $G$ Can this work? $2$ $1$ No, as $H+2$ and $D=1$ already. $4$ $2$ No, as $H=2$ already. $6$ $3$ Yes, this is possible.

So $B = 6$ and $G = 3$

$E - G = F$, therefore $E - 3 = F$.
The smallest possible value for $F$ is $4$, and this means that $E$ would be $7$.
$F$ cannot be any greater than $4$ because this would mean that $E > 7$ and this is not allowed.
Therefore $F = 4$ and $E = 7$

Now we can solve $A + H = E$.
This is $A + 2 = 7$, so $A = 5$.

Here are the solved equations:

 $A + C = A$ $5 + 0 = 5$ $F \times D = F$ $4 \times1 = 4$ $B - G = G$ $6 - 3 = 3$ $A + H = E$ $5 + 2 = 7$ $B / H = G$ $6 / 2 = 3$ $E - G = H$ $7 - 3 = 4$

And all the values of $A$-$H$:

 $A$ $5$ $B$ $6$ $C$ $0$ $D$ $1$ $E$ $7$ $F$ $4$ $G$ $3$ $H$ $2$