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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 |
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...