In the following question two positions of the same dice are given. Find the…

Concept: On a cube, any two adjacent faces meet along a shared edge, and that edge has two end-vertices. A different third face touches the pair at each end-vertex — and these two possible third faces are always opposite each other. So if two separate positions of the same die show an identical pair of adjacent faces but a different third face each time, those two third faces must be the ones opposite each other.

Application:

1. In the first position, the visible faces are 1 (top), 3 (front) and 2 (right); here 2 and 3 are adjacent to each other, with 1 as the third face at their shared vertex.

2. In the second position, the visible faces are 2 (top), 3 (front) and 6 (right); the same pair, 2 and 3, is adjacent again, but this time the third face at their shared vertex is 6.

3. Both positions show the identical adjacent pair (2, 3) with a different third face each time, so by the rule above, the two third faces — 1 and 6 — must be opposite each other.

Cross-check:

• 2 and 3 are each directly adjacent to 6 in the second position, so neither of them can be opposite 6.

• 4 and 5 never appear in either position, so the shared-edge rule cannot link either of them to 6; only 1 satisfies the rule.

• Hence the face opposite the face with 6 dots carries 1 dot.