AMC 10 · 2006 · #22
Grade 8 number-theoryPick an answer.
The question asks for a minimum, so Tool #14 (Extreme Principle) fixes the shape of the whole solution: prove a floor no triple can sink below, then exhibit a triple standing exactly on it. Both halves are mandatory, and dropping the first is the usual way this problem goes wrong. Tool #16 (Change Focus) makes any counting possible: chasing factors of 10 directly is hopeless, but 10 = 2 · 5 and factorials are far richer in twos than in fives, so n is nothing but a count of fives. Tool #7 (Identify Subproblems) counts those fives one power at a time — multiples of 5, then 25, then 125, then 625. Tool #15 (Organize Information in More Ways) is the move that unlocks the optimization: the twelve resulting counts get grouped by denominator instead of by variable, which is the only arrangement in which the constraint a + b + c = 2006 can be applied at all. Tool #4 (Introduce a Variable) names the remainders and proves the single inequality each group obeys; that inequality is the load-bearing claim of the solution, so it is proved from division with remainder rather than read off small cases. Tool #11 (Work Backwards) then runs the equality condition backwards to see what a, b, c must look like for all four groups to be tight simultaneously, and Tool #6 (Guess and Check) confirms the resulting triple by direct computation.
Count fives, not tens
Only the fives need counting, since they are scarcer.
Every trailing zero needs a two and a five, and a factorial is drowning in twos, so the fives are what run out first.
Every trailing zero needs a two and a five, and a factorial is drowning in twos, so the fives run out first.
▸ Why?
Each number has one prime recipe, so a trailing zero is exactly one pairing of a two with a five.
▸ Why?
Since the twos are always the more plentiful of the two, the smaller supply is what caps the count.
Count the fives inside one factorial
A standard formula counts them inside one factorial.
A number holding three fives gets picked up once by 5, once by 25 and once by 125, so counting multiples of each power tallies every five exactly once.
4.OA.B.4Identify SubproblemsGroup the twelve terms by denominator
Grouping by denominator lines the terms into four rows.
The only fact we own is that a, b and c add to 2006, and that fact becomes usable only once terms sharing a divisor stand side by side.
7.EE.A.2Organize Information In More WaysOne inequality every row obeys
Each row loses at most two to rounding.
Cutting 2006 into three pieces can waste at most two whole groups of size k, because each piece leaves behind less than one full group.
4.NBT.B.6Introduce A VariableAdd the four rows for a hard floor
Adding the rows gives a hard floor of 492.
Every row of the tally is anchored to 2006 and can sag at most two units below it, so four rows can sag at most eight.
7.EE.B.4Extreme PrincipleWork backwards to a tight triple
Working backwards finds a triple that hits every equality.
Sitting one step below 625 makes a number maximally wasteful at every power of five at once, which is exactly what tightness in all four rows requires.
6.EE.A.1Work BackwardsVerify the triple and read off n
Checking it confirms 492, choice (B).
A proven floor and a working example that agree leave no gap between them, so the minimum is nailed down rather than guessed.
4.NBT.B.6Guess And CheckTrailing zeros are really a count of fives, and splitting 2006 into three parts can waste at most two fives at each power of five — so prove the waste can never exceed eight, then pick numbers like 624 that waste the maximum at every power at once.
- Count fives, not tens
- Count the fives inside one factorial
- Group the twelve terms by denominator
- One inequality every row obeys
- Add the four rows for a hard floor
- Work backwards to a tight triple
- Verify the triple and read off n