Maths Olympiad Prep

Track / Stage 5 / 262 of 400 #862 of 1964

Problem 862

AIME late
Combinatorics Difficulty 5.6 Find the answer

B3. On each of the twelve edges of a cube, we write the number 1 or -1. We then multiply the four numbers on the edges of each face of the cube and write the result on that face. Finally, we add up the eighteen written numbers.

What is the smallest (most negative) result we can get this way? In the figure, you see an example of such a cube. The numbers on the back of the cube are not visible here.

!

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

3. -12

First, we show that the result is always -12 or more. If we write a -1 on each of the twelve edges, we get a 1 in each face. In that case, the result is 12+6=6-12+6=-6. For every -1 on an edge that we change to a 1, at most two faces change from a 1 to a -1. The result thus becomes at most 2 lower (1+1+1(-1+1+1 becomes 111)1-1-1). To go below -12, we would need to write a 1 on at least 4 edges. However, in that case, the result is

!
at least (48)6=10(4-8)-6=-10. The result will therefore never be lower than -12.

Finally, we show that you can indeed get -12 as a result. To do this, we write a -1 on each edge, except for the three edges indicated in the figure. On each face, exactly three edges have a -1. Therefore, each face has a -1. The result is thus (39)6=12(3-9)-6=-12.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.