Reasoning · Grade 5-2 Dice

Problem

Complete a die from several views

One die, whose opposite faces total 7, is drawn three times from the same angle. Top, front and right read 1, 2, 3 in the first, A, 2, 6 in the second, and 1, 3, B in the third. Find the pips opposite A and opposite B.
A B
Your answer
How to solve
Strategy Visualize Spatial Relationships — The 7-rule alone is not enough: it narrows face A down to just two candidates and then stops, because 3 and 4 are opposite each other and the rule cannot tell them apart. What separates them is the direction the die is turned, so the plan is to fill in the hidden faces of the first die, which fixes the whole die, and then notice that the second and third pictures each keep ONE face in its original place. That means each of them is the first die spun about that face, and spinning about a face sends the other four faces round a fixed ring in a fixed order. Reading a face off a ring is much safer than trying to picture a rotation, and the whole thing can be checked in ten seconds with a real die in your hand.
1STEP 1

Fill in the hidden half of the first die

Opposites add to 7, filling in the hidden half.

7-1=6, 7-2=5, 7-3=4
2STEP 2

Narrow face A down to two possibilities

Face A narrows to 3 or 4.

{1,6} left-right, {2,5} front-back → A∈{3,4}
3STEP 3

Notice that the second die is the first die spun about the face marked 2

The second picture spins about the face marked 2.

4STEP 4

Read face A off the ring

Round the ring, A is 3.

ring (top→right→bottom→left): 1→ 3→ 6→ 4→ 1; right=6→ top=3; 7-3=4
5STEP 5

Do the same for the third die, spinning about the face marked 1

The same trick makes B 5.

ring (front-left→right→back-right→back-left): 2→ 3→ 5→ 4→ 2; front-left=3→ right=5; 7-5=2
6STEP 6

Check both answers on a real die

The opposite faces are 4 and 2.

Answer
4, 2
7 − 3 = 4, 7 − 5 = 2
Every number involved is a pip count between 1 and 6, and both answers, 4 and 2, are in that range. The pieces also fit together with no clashes: the second die then reads top 3, front-left 2, right 6, with hidden faces 4, 5, 1 — all six numbers used exactly once — and the third die reads top 1, front-left 3, right 5, with hidden faces 6, 4, 2 — again all six used exactly once. Each answer is the 7-complement of the face I deduced (3+4=7 and 5+2=7), and A came out as 3, one of the two candidates the elimination step had already allowed, which is a genuine independent check on the ring argument. Finally the deduction is forced, not merely consistent: with the first die fixed, only one value of A makes the second picture possible, and only one value of B makes the third picture possible.
Takeaway

When two pictures of a die share a face, the die was just spun about that face — so the other four faces are still in the same ring, and you can read the hidden one straight off.

  • Fill in the hidden half of the first die
  • Narrow face A down to two possibilities
  • Notice that the second die is the first die spun about the face marked 2
  • Read face A off the ring
  • Do the same for the third die, spinning about the face marked 1
  • Check both answers on a real die