Problem 404 worked answer

404: The Address That Survived

Designed for Grade 5 · Uses Middle Grade 3 math · About 5–13 minutes

Complete to collectThe Missing JewelLevel 3

At a glance

  • Part 1: The vanished address is MATH. The mismatching letters are P in PATH, O in MOTH, S in MASH, and E in MATE.
  • Part 2: Yes. Every set of three suggestions still forces MATH.
  • Part 3: PATH and MATE allow exactly MATH and PATE. Rook must keep 3 suggestions to guarantee recovery no matter which suggestions remain.
  • Part 4: Answers vary. BATH is one example; its first letter differs from MATH.

1. Recover the address

Compare the four suggestions by position:

PositionLetters in the four suggestionsLetter forced by the report
1P, M, M, MM
2A, O, A, AA
3T, T, S, TT
4H, H, H, EH

This gives MATH. It differs from every suggestion in exactly one position:

  • PATH has the wrong letter P in position 1;
  • MOTH has the wrong letter O in position 2;
  • MASH has the wrong letter S in position 3; and
  • MATE has the wrong letter E in position 4.

So MATH satisfies all four clues. The four circled mismatch letters happen to spell POSE. That is an Easter egg, not a fact needed for the solution.

2. Prove that any three suggestions are enough

Take any two suggestions from one of the three-row sets. Because their wrong letters occur in different positions, the two suggestions differ in exactly two positions. A four-letter address that is one position away from both can only combine those two differing positions in one of two ways:

  1. the combination that produces MATH; or
  2. the crossed combination that keeps both wrong letters.

The third suggestion has its wrong letter in a different position. It rejects the crossed combination, which is then three positions away from that third suggestion. The third suggestion still accepts MATH.

Here is the full check for every possible lost row:

Lost suggestionFirst two retained cluesOther common one-away stringThird retained clueWhy the other string fails
PATHMOTH and MASHMOSHMATEMOSH is three positions from MATE
MOTHPATH and MASHPASHMATEPASH is three positions from MATE
MASHPATH and MOTHPOTHMATEPOTH is three positions from MATE
MATEPATH and MOTHPOTHMASHPOTH is three positions from MASH

In every row, MATH remains one position from all three retained suggestions and the only competing string is rejected. Therefore Rook is correct no matter which one suggestion is lost.

3. Find the smallest guaranteed clue count

PATH and MATE agree in positions 2 and 3. Any address that is one position away from both must keep those shared letters, A and T. The words differ in positions 1 and 4:

P A T H
M A T E

To remain one position from each word, the address must take one of those two differing positions from PATH and the other from MATE. There are exactly two ways:

  • MATH takes M from MATE and H from PATH;
  • PATE takes P from PATH and E from MATE.

No third address works. Changing a shared position would create an additional mismatch, and taking both differing positions from the same listed word would reproduce that word rather than remain one position from it.

Part 2 proves that every surviving set of three suggestions forces MATH. The PATH–MATE example proves that keeping only two suggestions does not guarantee a unique address: those two permit both MATH and PATE. Therefore the fewest suggestions Rook must keep to guarantee recovery, no matter which suggestions remain, is 3.

4. Construct another suggestion

Any familiar four-letter word other than the four listed suggestions is valid when it differs from MATH in exactly one position. Examples include:

  • BATH — position 1 changes from M to B;
  • LATH — position 1 changes from M to L;
  • MATS — position 4 changes from H to S;
  • MYTH — position 2 changes from A to Y; and
  • MACH — position 3 changes from T to C.

A response is not valid if it changes two or more positions, repeats PATH, MOTH, MASH, or MATE, or simply writes MATH unchanged.

Why the proof is complete

The reconstruction checks every supplied row. The any-three table covers all four possible lost suggestions. The PATH–MATE argument lists both and only the possible recombinations, so it supplies a valid counterexample to a two-clue guarantee. Together those arguments establish both sides of the minimum claim: three always suffice, while two need not.

Check the information-recovery connection

The worksheet compares strings by the number of aligned positions where they differ. That count is their Hamming distance. Rook's report supplies redundant evidence: even after any one row is lost, the remaining three rows still recover the same address. Error-correcting systems use more structured forms of the same broad idea—extra information can make lost or damaged information recoverable.

Technical fit and rating

MJ 3.5 · C3 · W2 Stretch challenge · Moderate workload

Discrete mathematics and logic

How the rating works →

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