AMC 10 · 2003 · #13
Easy mode Grade 2Here is a rule: take a number and add up its digits. For example, the digits of 8 add to 8, and the digits of 123 add to 1+2+3=6. Now use this rule twice: first add the digits of a number, then add the digits of that result. For how many two-digit numbers does this final total equal 3?
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The symbol $\clubsuit(x)$ means: add up the digits of $x$. So $\clubsuit(\clubsuit(x))$ means do it twice — first add the digits of $x$, then add the digits of that result. Among the two-digit numbers ($10$ through $99$), count how many make this double digit-sum equal to $3$.
Givens: $\clubsuit(x)$ is the sum of the digits of $x$; e.g. $\clubsuit(8)=8$ and $\clubsuit(123)=6$.; $x$ ranges over the two-digit numbers, $10$ through $99$.; We want $\clubsuit(\clubsuit(x)) = 3$.; Answer choices: (A) $3$, (B) $4$, (C) $6$, (D) $9$, (E) $10$.
Unknowns: How many two-digit values of $x$ satisfy $\clubsuit(\clubsuit(x)) = 3$.
Understand
Restated: The symbol $\clubsuit(x)$ means: add up the digits of $x$. So $\clubsuit(\clubsuit(x))$ means do it twice — first add the digits of $x$, then add the digits of that result. Among the two-digit numbers ($10$ through $99$), count how many make this double digit-sum equal to $3$.
Givens: $\clubsuit(x)$ is the sum of the digits of $x$; e.g. $\clubsuit(8)=8$ and $\clubsuit(123)=6$.; $x$ ranges over the two-digit numbers, $10$ through $99$.; We want $\clubsuit(\clubsuit(x)) = 3$.; Answer choices: (A) $3$, (B) $4$, (C) $6$, (D) $9$, (E) $10$.
Plan
Primary tool: #11 Work Backwards
Secondary: #14 Extreme Principle, #2 Make a Systematic List, #7 Identify Subproblems
The condition is stacked: the outer $\clubsuit$ acts on the inner one. So work backwards (Tool #11) — name the inner digit sum $s = \clubsuit(x)$ and first ask which values of $s$ make $\clubsuit(s) = 3$. To keep that search finite, pin down the range of $s$ with a boundary check (Tool #14): a two-digit number's digits can sum to at most $9+9=18$, so $s$ lives in $1$ to $18$. That turns the outer condition into just two allowed inner sums, $s=3$ and $s=12$. Then the problem splits into two independent counting subproblems (Tool #7), each solved by a short systematic list (Tool #2): count two-digit numbers whose digits add to $3$, and those whose digits add to $12$. Add the two counts.
Execute — Answer: E
1.NBT.B.2 Step 1 Name the inner sum and bound it
- Let $s = \clubsuit(x)$ be the first digit sum.
- A two-digit number has a tens digit from $1$ to $9$ and a ones digit from $0$ to $9$, so the smallest its digits can add to is $1+0 = 1$ and the largest is $9+9 = 18$.
- Therefore $s$ is somewhere between $1$ and $18$.
- This finite range is what makes the outer condition easy to handle.
💡 The biggest a two-digit digit-sum can reach is $9+9=18$, so the inner value can never leave $1$–$18$.
2.OA.B.2 Step 2 Find which inner sums work
- The condition is $\clubsuit(s) = 3$: the digits of $s$ must themselves add to $3$.
- Check the range $1$ to $18$.
- A one-digit $s$ works only when $s = 3$ (since its single digit is $3$).
- A two-digit $s$ (that is, $10$ through $18$) works only when its digits add to $3$, which means $s = 12$.
- No other value in $1$–$18$ qualifies.
- So the inner sum must be $s = 3$ or $s = 12$.
💡 Only $3$ and $12$ have digits that add to $3$ once you stay at or below $18$.
1.NBT.B.2 Step 3 Count two-digit numbers with digit sum 3
- Now count the two-digit numbers whose digits add to $3$.
- Write $x$ as tens digit $a$ (from $1$ to $9$) and ones digit $b$ (from $0$ to $9$) with $a + b = 3$.
- Running the tens digit up: $a=1,b=2$ gives $12$; $a=2,b=1$ gives $21$; $a=3,b=0$ gives $30$.
- If $a=4$ then $b=-1$, impossible, so the list stops.
- That is $3$ numbers.
💡 For a small target sum the tens digit can only be $1$, $2$, or $3$, so there are just three numbers.
1.NBT.B.2 Step 4 Count two-digit numbers with digit sum 12
- Next count the two-digit numbers whose digits add to $12$, again with tens digit $a$ from $1$ to $9$ and ones digit $b$ from $0$ to $9$, $a + b = 12$.
- The ones digit $b = 12 - a$ must stay between $0$ and $9$, which forces $a$ from $3$ up to $9$: $39, 48, 57, 66, 75, 84, 93$.
- That is $7$ numbers.
💡 The ones digit can be at most $9$, so the tens digit must be at least $3$, leaving exactly the seven values $3$ through $9$.
2.OA.A.1 Step 5 Add the two counts
- The two cases $s=3$ and $s=12$ never overlap — a number's digit sum is a single value, so no $x$ is counted twice.
- Add the counts: $3 + 7 = 10$ two-digit numbers in all.
- Among the choices, $10$ is option (E).
💡 The two digit-sum cases are separate, so the totals simply add.
1.NBT.B.2 Let $s = \clubsuit(x)$ be the first digit sum. A two-digit number has a tens dig 2.OA.B.2 The condition is $\clubsuit(s) = 3$: the digits of $s$ must themselves add to $3 1.NBT.B.2 Now count the two-digit numbers whose digits add to $3$. Write $x$ as tens digit 1.NBT.B.2 Next count the two-digit numbers whose digits add to $12$, again with tens digit 2.OA.A.1 The two cases $s=3$ and $s=12$ never overlap — a number's digit sum is a single Review
Reasonableness: The count $10$ is comfortably inside what's possible: there are $90$ two-digit numbers, and requiring a specific double digit-sum should keep only a small handful, so $10$ is reasonable. Spot-check the two edge members: $12 \to \clubsuit(12)=3 \to \clubsuit(3)=3$ ✓, and $93 \to \clubsuit(93)=12 \to \clubsuit(12)=3$ ✓. Both lists are also self-consistent: digit sums $3$ and $12$ are exactly the values in $1$–$18$ whose digits add to $3$, so nothing was missed and nothing extra slipped in.
Alternative: Skip naming $s$ and just scan by first digit sum. Group the two-digit numbers by their digit sum $1,2,\dots,18$; the outer condition $\clubsuit(s)=3$ selects only the groups with sum $3$ or $12$. The 'sum $= 3$' group has $3$ members and the 'sum $= 12$' group has $7$, giving $3+7=10$ again. This is the same split, organized as a table of digit-sum groups (Tool #15, Organize Information).
CCSS standards used (min grade 2)
1.NBT.B.2Understand that the two digits of a two-digit number represent tens and ones (Treating $x$ as a tens digit plus a ones digit to bound the digit sum ($1$–$18$) and to list all numbers with a given digit sum.)2.OA.B.2Fluently add and subtract within 20 using mental strategies (Adding digits to test $\clubsuit(s)=3$ and pin the inner sum to $s=3$ or $s=12$.)2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Combining the two non-overlapping counts, $3 + 7 = 10$.)
⭐ Work from the outside in: figure out what the first digit sum must be ($3$ or $12$), then list the two-digit numbers that hit each and add the counts.
⭐ Work from the outside in: figure out what the first digit sum must be ($3$ or $12$), then list the two-digit numbers that hit each and add the counts.
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