Push a straight rod from one corner of a box of unit cubes to the opposite corner. It clips some cubes and misses others — and counting them one by one is hopeless for a big box, so the trick is to count the walls it crosses instead.
The rod enters a new little cube every time it crosses an inside wall, so a first guess is just how many walls there are: a + b + c. But that over-counts, because sometimes the rod crosses two walls at the very same point — it slips through the shared edge of two cubes and only enters ONE new cube, not two. How often that happens for each pair of directions is exactly the greatest common divisor of those two edge lengths, so the three pairwise GCDs come off the total.
Why this sheet works
Take one off too many, though: where all three walls meet at a single lattice point, the correction was applied twice, so the GCD of all three edges is added back. The whole count is a + b + c − gcd(a,b) − gcd(b,c) − gcd(c,a) + gcd(a,b,c) — inclusion and exclusion made physical, and a first glimpse of why the humble GCD is really about geometry.
A rod is pushed corner to corner through a block of unit cubes. Count the ones it passes through, and show why a + b + c is too many.
1A solid block is built from unit cubes, 3 along one edge, 7 along the next and 8 along the last. A straight rod is pushed from one corner right through to the opposite corner.How many of the little cubes does it pass through?Why
2A cuboid of sugar is 2 small cubes wide, 3 deep and 4 tall. A thin skewer is run from a bottom corner straight to the far top corner.How many of the small cubes does it pass through?Why
Cubes a diagonal piercesEdges up to 8
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A rod is pushed corner to corner through a block of unit cubes. Count the ones it passes through, and show why a + b + c is too many.
3A 2 × 6 × 6 box is packed exactly with unit cubes. A laser is aimed from one corner to the corner diagonally opposite, passing straight through the block.How many of the unit cubes does it pass through?Why
4A 2 × 7 × 8 box is packed exactly with unit cubes. A laser is aimed from one corner to the corner diagonally opposite, passing straight through the block.How many of the unit cubes does it pass through?Why
Cubes a diagonal pierces · Edges up to 8Answer key
1. Corner to corner, the rod enters a new little cube at every wall it crosses — 18 of them if none lined up, which is the count to beat. But two walls meeting on one line are crossed together: that happens gcd(3,7) = 1, gcd(7,8) = 1 and gcd(8,3) = 1 times, 3 in all, each an over-count of one. Taking them off: 18 − 3 = 15. All three walls meet together gcd(3,7,8) = 1 times, and that was subtracted once too often, so add it back: 15 + 1 = 16.2. Corner to corner, the rod enters a new small cube at every wall it crosses — 9 of them if none lined up, which is the count to beat. But two walls meeting on one line are crossed together: that happens gcd(2,3) = 1, gcd(3,4) = 1 and gcd(4,2) = 2 times, 4 in all, each an over-count of one. Taking them off: 9 − 4 = 5. All three walls meet together gcd(2,3,4) = 1 times, and that was subtracted once too often, so add it back: 5 + 1 = 6.3. Corner to corner, the rod enters a new unit cube at every wall it crosses — 14 of them if none lined up, which is the count to beat. But two walls meeting on one line are crossed together: that happens gcd(2,6) = 2, gcd(6,6) = 6 and gcd(6,2) = 2 times, 10 in all, each an over-count of one. Taking them off: 14 − 10 = 4. All three walls meet together gcd(2,6,6) = 2 times, and that was subtracted once too often, so add it back: 4 + 2 = 6.4. Corner to corner, the rod enters a new unit cube at every wall it crosses — 17 of them if none lined up, which is the count to beat. But two walls meeting on one line are crossed together: that happens gcd(2,7) = 1, gcd(7,8) = 1 and gcd(8,2) = 2 times, 4 in all, each an over-count of one. Taking them off: 17 − 4 = 13. All three walls meet together gcd(2,7,8) = 1 times, and that was subtracted once too often, so add it back: 13 + 1 = 14.
