AMC 10 · 2007 · #22
Easy mode Grade 4For any whole number n bigger than 0, let S(n) be the sum of its digits. For example, S(2007)=2+0+0+7=9. Applying S again, S(S(n)) is the digit sum of that result. For how many values of n is n+S(n)+S(S(n))=2007?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: For a positive integer $n$, let $S(n)$ be the sum of its digits, and apply $S$ again to get $S(S(n))$. Count how many $n$ make $n + S(n) + S(S(n))$ exactly equal to $2007$.
Givens: $S(n)$ is the sum of the digits of $n$; $S(S(n))$ is the digit sum of that digit sum; The equation to satisfy is $n + S(n) + S(S(n)) = 2007$; $n$ ranges over positive integers; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: The number of positive integers $n$ that satisfy the equation
Understand
Restated: For a positive integer $n$, let $S(n)$ be the sum of its digits, and apply $S$ again to get $S(S(n))$. Count how many $n$ make $n + S(n) + S(S(n))$ exactly equal to $2007$.
Givens: $S(n)$ is the sum of the digits of $n$; $S(S(n))$ is the digit sum of that digit sum; The equation to satisfy is $n + S(n) + S(S(n)) = 2007$; $n$ ranges over positive integers; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #14 Extreme Principle
Secondary: #16 Change Focus / Count the Complement, #2 Make a Systematic List
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.
Execute — Answer: D
4.NBT.B.4 Step 1 Trap $n$ in a small window
- Since $S(n)$ and $S(S(n))$ are both positive, $n < 2007$.
- Now bound how big those two terms can be.
- For any $n$ below $2007$, the digits are at most $1,9,9,9$, so $S(n) \le 28$.
- Then $S(n)$ is at most $28$, whose digit sum is at most $10$, so $S(S(n)) \le 10$.
- Hence $S(n) + S(S(n)) \le 38$.
- Rearranging the equation, $n = 2007 - S(n) - S(S(n)) \ge 2007 - 38 = 1969$.
- So every solution satisfies $1969 \le n \le 2006$.
💡 Digit sums stay tiny, so $n$ can only be a short step below $2007$.
4.NBT.B.6 Step 2 Keep only multiples of $3$
- A number leaves the same remainder as its digit sum when divided by $9$, because $10, 100, 1000$ are each one more than a multiple of $9$.
- So $n$, $S(n)$, and $S(S(n))$ all leave the same remainder $r$ when divided by $9$.
- Their total then leaves remainder $r + r + r = 3r$.
- We need the total to be $2007$, and $2007 = 9 \times 223$ leaves remainder $0$.
- So $3r$ must be a multiple of $9$, which means $r$ is a multiple of $3$.
- A number whose remainder mod $9$ is a multiple of $3$ is exactly a multiple of $3$.
- So $n$ must be a multiple of $3$.
💡 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 $3$.
4.NBT.A.2 Step 3 Test each surviving candidate
- List the multiples of $3$ from $1971$ up to $2004$ and compute $n + S(n) + S(S(n))$ for each.
- $1971 \to 1971+18+9 = 1998$.
- $1974 \to 1974+21+3 = 1998$.
- $1977 \to 1977+24+6 = 2007$.
- $1980 \to 1980+18+9 = 2007$.
- $1983 \to 1983+21+3 = 2007$.
- $1986 \to 1986+24+6 = 2016$.
- $1989 \to 1989+27+9 = 2025$.
- $1992 \to 1992+21+3 = 2016$.
- $1995 \to 1995+24+6 = 2025$.
- $1998 \to 1998+27+9 = 2034$.
- $2001 \to 2001+3+3 = 2007$.
- $2004 \to 2004+6+6 = 2016$.
💡 With the window this small, a direct place-by-place check of each multiple of $3$ settles the count for sure.
4.OA.B.4 Step 4 Count the hits
- Exactly four candidates hit $2007$: $n = 1977$, $1980$, $1983$, and $2001$.
- No others in the window work, and the window contains every possible solution.
- So there are $4$ values of $n$.
- The answer is (D).
💡 Four numbers survive every filter and every check, so the count is four.
4.NBT.B.4 Since $S(n)$ and $S(S(n))$ are both positive, $n < 2007$. Now bound how big thos 4.NBT.B.6 A number leaves the same remainder as its digit sum when divided by $9$, because 4.NBT.A.2 List the multiples of $3$ from $1971$ up to $2004$ and compute $n + S(n) + S(S(n 4.OA.B.4 Exactly four candidates hit $2007$: $n = 1977$, $1980$, $1983$, and $2001$. No o Review
Reasonableness: Spot-check the four winners by hand. $1977$: digits sum to $1+9+7+7 = 24$, and $2+4 = 6$, giving $1977+24+6 = 2007$. $2001$: digits sum to $3$, and $S(3)=3$, giving $2001+3+3 = 2007$. Both check out. The four solutions land in two natural clusters ($1977$–$1983$ and the lone $2001$), consistent with the digit sums jumping around as $n$ crosses from the $1900$s into the $2000$s. Four is one of the offered choices and matches the multiple-of-$3$ pattern, so (D) is consistent.
Alternative: Skip the mod-$9$ shortcut and brute-force the window directly. Since $1969 \le n \le 2006$ is only $38$ numbers, one can simply compute $n + S(n) + S(S(n))$ for every $n$ in that range. The same four values — $1977, 1980, 1983, 2001$ — turn up, confirming the count without the divisibility argument; the mod-$9$ filter just saves about two thirds of the arithmetic.
CCSS standards used (min grade 4)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Bounding $n$ between $1969$ and $2006$ from $2007 - S(n) - S(S(n))$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Using remainders mod $9$ and the digit-sum rule to force $n$ to be a multiple of $3$.)4.NBT.A.2Read and write multi-digit whole numbers using base-ten numerals and expanded form (Computing each digit sum $S(n)$ and $S(S(n))$ for the candidate numbers.)4.OA.B.4Find all factor pairs for a whole number and recognize multiples (Restricting candidates to multiples of $3$ and counting the four that work.)
⭐ Because digit sums are tiny, $n$ must sit just below $2007$; checking that short list of multiples of $3$ leaves exactly four winners.
⭐ Because digit sums are tiny, $n$ must sit just below $2007$; checking that short list of multiples of $3$ leaves exactly four winners.
More like this
Same archetype — closest grade level first.