Casework

Try one possible hiding place. What would each card say about that place?

Answer: The badge is in B. A and C would each make two cards true, breaking the promise.

Test each case

Suppose the badge is in A. Follow what that would mean for all three cards. Then start afresh with B, and again with C. Supposing a place does not let us change a clue to make it fit.

Check B: “not in A” is true, “in A” is false, and “not in B” is false. Exactly one is true. A, B and C are every possible hiding place, and only B passes.

A shorter route is valid too. The first two cards say exact opposites, so exactly one of them is already true. Card three must therefore be false: the badge is in B. The case table makes the same reasoning visible one possibility at a time.

Answer: No. Use 3 rows of 6 cards. Two equal rows would need 9 cards each, exceeding the limit of 8.

The attempt used the wrong number of cards per row. It did not show that every three-row arrangement fails. Once we choose three equal rows, the total forces \(18\div3=6\) cards in each.

Check: \(3\times6=18\), the rows are equal and \(6\leq8\). In the two-row case, using all 18 forces \(18\div2=9\). There is no different equal two-row size that uses all the cards.

Reject a whole case only when its conditions cannot all hold. A failed attempt may leave another choice inside that same case.

Answer: \(41\times7=287\). This is the only winning assignment of tiles to those roles.

Fix one choice

There are three possible multiplier tiles. Hold one fixed. Then the two remaining digits can make just two two-digit numbers.

The larger remaining digit belongs in the tens place. For each fixed positive multiplier, that gives the best product in its case.

Now compare the three case winners. The greatest is 287.

Why is the search complete? Every legal display has multiplier 1, 4 or 7. In each case we kept its larger product. No discarded arrangement can beat its own case winner, and no other multiplier is available.

Check: 41 uses 4 and 1; its multiplier uses the remaining 7. Each tile appears once. The other case winners are 74 and 284, both smaller. Calculating all six legal products would also work; fixing one choice reduces the comparison to three justified winners.

Which yes-or-no feature could divide the packs into two lists that cannot overlap? Within a list, what choices remain?

Answer: 11 packs. With Train: Train–Star–Moon, Train–Star–Leaf, Train–Star–Wave, Train–Star–Sun, Train–Moon–Wave and Train–Leaf–Wave. Without Train: Star–Moon–Wave, Star–Moon–Sun, Star–Leaf–Wave, Star–Leaf–Sun and Star–Wave–Sun.

Casework uses Train present and Train absent. Every pack has exactly one of those states. The second case forces Star; the first still allows it. This removes overlap without discarding packs that contain both.

Make an organized list

Keep the design order Train, Star, Moon, Leaf, Wave, Sun. In each case, fix the earlier remaining design and try every later partner. Cross out a pair if it uses Moon with Leaf, or uses Sun without Star. A rule inside a case can reject a pair; it cannot justify skipping other pairs.

Check: six Train-present packs and five Train-absent packs give \(6+5=11\). Every listed pack has three different designs, includes Train or Star, avoids Moon with Leaf and includes Star whenever Sun occurs. The cases cannot overlap, and the ordered partner lists include every possible pair inside each case.

A direct list of all twenty three-design selections and a check of each is valid too. The chosen split makes the Star requirement automatic in one whole part and keeps the no-repeat argument visible.

Answer: Plan C, with 22 Sun children and 26 Moon children: 48 children in 12 complete rows. A fails the counter total; B fails the seating condition.

Casework gives three possible plans to investigate. A plan only survives if its group sizes, counter total and seating all fit.

Use even and odd

Plan A gives every child a pair of counters. Those pairs total an even number. Adding its one spare gives an odd number, whatever the group sizes. It cannot make the even total 144. This rejects the whole plan without trying headcounts.

Use a variable

For B and C, the group counts are unknown but cannot vary independently. Name the Sun count and preserve the four-child difference in every counter expression; checking unrelated guesses would repeatedly lose that link.

Write it with a variable

B and C give different amounts to the two groups. We need to keep both unknown group sizes connected. Let \(n\) mean the number of Sun children. Then \(n+4\) is the number of Moon children. The same letter keeps the same meaning throughout.

In Plan B, \(3n\) counts the Sun counters, \(5(n+4)\) counts the Moon counters, and 4 counts the spares. The whole Moon count is multiplied by five.

All counters must be accounted for, so Plan B would require:

That would mean 15 Sun and 19 Moon children: 34 altogether. The counters fit, but the seats do not. Eight complete rows hold 32, leaving two children. Reject B using the original seating clue.

For Plan C the counter equation changes, while the meaning of \(n\) stays fixed:

Now Sun has 22 children and Moon has \(22+4=26\). That makes \(22+26=48\), which fills \(48\div4=12\) rows.

Check C: the groups are nonempty whole counts and \(26-22=4\). The counters total:

Finally, \(12\times4=48\) seats hold all 48 children. Every clue is satisfied.

Why only this record? A always has the wrong parity. B’s equation forces its only possible Sun count, 15, which fails seating. C’s equation forces 22, which works. In each equation increasing the Sun count increases the total, so a second Sun count cannot give 144. The three plans cover all allowed choices.

How the tools fit: Casework keeps the competing plans separate. Even/odd reasoning excludes all of A at once. Use a variable connects the two group sizes so we can test B and C completely. You could reason in words about matched Sun–Moon pairs instead; the letter is useful bookkeeping for the relationships, rather than a new rule to memorise.

Revisit Use even and odd to test a necessary condition. Passing it alone does not prove every clue fits.