brainteaser

Brainteasers #2

$A, B, C, D, E, F, G, H, I$ are the nine integers from one to nine (not necessarily in order). They satisfy the following constraints:

$A + B +C +D = 20$
$B+C+D+E+F = 20$
$D+E+F+G+H = 20$
$F+G+H+I=20$

What values are taken by each of A to I?

Answer below:

Since $A, B, C, D$ and $F, G, H, I$ sum to 40, $E$ must be 5.

If E is 5, subtracting equation 1 with equation 2, A - 5 - F = 0; subtracting equation 3 with equation 4, D +5 - I = 0. So, any combinations that satified these two equations can be the answer. For example, A=9, F=4, D=2, I=7.

With this, only B, C, G, H are left. Subtracting equation 2 with equation 3, B + C = G + H. There are again many combinations depending on what you chose for A, F, D, I. If we used the above numbers, what’s left is 1, 3, 6, 8. So, one answer can be B=1, C=8, G=6, H=3.

Brainteasers #2

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