Case 06 · integer arithmetic · Medium · About 6–10 minutes
An airship's freight clerk has three kinds of sealed crate. Their sizes look different, but their gross masses are standardized. A smudged manifest preserves the total count and total mass, plus one small packing rule.
Recover the number of each kind without opening any crates.
Every crate is exactly one of these types:
- A light crate has a gross mass of 3 kg.
- A medium crate has a gross mass of 5 kg.
- A heavy crate has a gross mass of 8 kg.
The manifest gives three facts:
- There are exactly 17 crates altogether.
- Their combined gross mass is exactly 94 kg.
- The number of medium crates is even.
There is at least one crate of each type. Counts must be whole numbers. Gross mass already includes the container, so do not add packaging weight.
How many light, medium, and heavy crates are aboard?
You may use algebra or a short table. If you make a table, use one row for each possible heavy-crate count. Work out what mass and crate count remain for the other two types. Remember that a matching total mass is only part of the test; the even-count rule must also hold.
Hint 1 - reveal when you want a nudge
Imagine all 17 crates weighed 3 kg. That would give 51 kg. Medium crates add 2 kg each above that baseline, and heavy crates add 5 kg each.
Hint 2 - reveal when you want a nudge
The extra mass is 43 kg, so 2m +5h =43. This forces the heavy-crate count to be odd.
Reveal the solution and explanation
The manifest contains 6 light, 4 medium, and 7 heavy crates.
Let m be the medium count and h the heavy count. The light count is 17 − m − h. Substituting into the mass total gives 51 + 2m + 5h = 94, or 2m + 5h = 43.
Because 43 is odd, h must be odd. It cannot exceed 8, since 9 heavy crates alone would exceed the available extra mass. Test h = 1, 3, 5, 7.
The corresponding medium counts are 19, 14, 9, 4. The first two leave no positive light count. The third gives three light crates but an odd medium count. Only h = 7, m = 4 survives. It leaves six light crates.
Check: 6 + 4 + 7 = 17, and 18 + 20 + 56 = 94. All types are present, and four is even. The short odd-number check covers every possible heavy count, proving uniqueness.
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