### Tweedle Dum and Tweedle Dee

Two brothers were left some money, amounting to an exact number of pounds, to divide between them. DEE undertook the division. "But your heap is larger than mine!" cried DUM...

### Fibs

The well known Fibonacci sequence is 1 ,1, 2, 3, 5, 8, 13, 21.... How many Fibonacci sequences can you find containing the number 196 as one of the terms?

### Seven Up

The number 27 is special because it is three times the sum of its digits 27 = 3 (2 + 7). Find some two digit numbers that are SEVEN times the sum of their digits (seven-up numbers)?

# Letter Land

##### Age 11 to 14 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 $A$-$H$:

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