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Not a Polite Question

When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square...

Whole Numbers Only

Can you work out how many of each kind of pencil this student bought?

Symmetricality

Add up all 5 equations given below. What do you notice? Solve the system and find the values of a, b, c , d and e. b + c + d + e = 4 a + c + d + e = 5 a + b + d + e = 1 a + b + c + e = 2 a + b + c + d = 0

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 $A$-$H$:

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