March 2015 - Six Boxes (Solution)

The answer is 84. Let Box 1 have the number a, Box 2 has b, Box 3 has c, etc. We have 6 equations:

a = e - d - f;
b = c + d - a;
c = a + 2d;
d = a - 3;
e = b + f - 1;
f = 2a - d.

Now, we can replace d by a - 3. This gives the following:

a = e - (a - 3) - f = e - a - f + 3;
b = c + (a - 3) - a = c - 3;
c = a + 2(a - 3) = 3a - 6;
e = b + f - 1;
f = 2a - (a - 3) = a + 3.

Now we can replace b by c - 3 and f by a + 3. This gives the following:

a = e - a - (a+3) + 3 = e - 2a;
c = 3a - 6;
e = (c - 3) + (a + 3) - 1 = c + a - 1.

Since a = e - 2a, it follows that e = 3a. This gives the following:

c = 3a - 6;
3a = c + a - 1.

Now, c = (c + a - 1) - 6 = c + (a - 7). Subtracting c from both sides, a - 7 = 0. Thus a = 7. Now, e = 3*7 = 21; c = 3*7 - 6 = 15; b = 15 - 3 = 12; f = 7 + 3 = 10; d = 7 - 3 = 4.

The least number is d, which is 4. The greatest number is e, which is 21. Thus the answer is 21*4 = 84.

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