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 $A$-$H$:

$A$ | $5$ |

$B$ | $6$ |

$C$ | $0$ |

$D$ | $1$ |

$E$ | $7$ |

$F$ | $4$ |

$G$ | $3$ |

$H$ | $2$ |