## 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; a = e - (a - 3) - f = e - a - f + 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; Since a = e - 2a, it follows that e = 3a. This gives the following:
c = 3a - 6; 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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