AMC 10 · 2007 · #22
Grade 4 number-theoryPick an answer.
There are infinitely many positive integers, so Tool #14 (Extreme Principle) first traps n in a tiny window: the two extra terms S(n)+S(S(n)) can never be large, so n must sit just below 2007. That leaves only a few dozen candidates. Tool #16 (Change Focus) then looks at the problem through remainders when dividing by 9; the digit-sum rule forces n to be a multiple of 3, cutting the candidates by two thirds. Finally Tool #2 (Make a Systematic List) walks the surviving multiples of 3 in order, computing n + S(n) + S(S(n)) for each and counting the hits.
Trap n in a small window
Bounding the extra terms traps the number in a short window.
Digit sums stay tiny, so n can only be a short step below 2007.
4.NBT.B.4Extreme PrincipleKeep only multiples of 3
A remainder rule keeps only the multiples of three.
Each term carries the same leftover mod 9, so their triple sum can only reach a multiple of 9 when n is a multiple of 3.
Each term leaves the same remainder as the number itself, so the triple sum reaches a multiple of nine only when the number does.
▸ Why?
A number and the sum of its digits differ by a multiple of nine, because every place value is one more than such a multiple.
▸ Why?
Three copies of the same remainder add to a multiple of nine only when that remainder is a multiple of three.
Test each surviving candidate
Testing the survivors finds the ones that hit the target.
With the window this small, a direct place-by-place check of each multiple of 3 settles the count for sure.
4.NBT.A.2Make A Systematic ListCount the hits
Counting them gives 4, choice (D).
Four numbers survive every filter and every check, so the count is four.
4.OA.B.4Make A Systematic ListBecause digit sums are tiny, n must sit just below 2007; checking that short list of multiples of 3 leaves exactly four winners.
- Trap n in a small window
- Keep only multiples of 3
- Test each surviving candidate
- Count the hits