AMC 10 · 2016 · #6
Easy mode Grade 4Laura adds two three-digit numbers. All six digits she uses are different from one another. Her answer is also a three-digit number, called S. Now add up the three digits of S. What is the smallest this digit total can be?
Pick an answer.
AMC 10 2016 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: Two three-digit numbers are added. The six digits used in the two numbers are all different. Their sum $S$ is also a three-digit number. Find the smallest possible value of the sum of the digits of $S$.
Givens: Two three-digit positive integers are added; All six digits across the two numbers are different from each other; The sum $S$ is a three-digit number; Answer choices: (A) $1$, (B) $4$, (C) $5$, (D) $15$, (E) $20$
Unknowns: The smallest possible digit sum of $S$
Understand
Restated: Two three-digit numbers are added. The six digits used in the two numbers are all different. Their sum $S$ is also a three-digit number. Find the smallest possible value of the sum of the digits of $S$.
Givens: Two three-digit positive integers are added; All six digits across the two numbers are different from each other; The sum $S$ is a three-digit number; Answer choices: (A) $1$, (B) $4$, (C) $5$, (D) $15$, (E) $20$
Plan
Primary tool: #14 Extreme Principle
Secondary: #3 Eliminate Possibilities, #6 Guess and Check
The question asks for a minimum, so this is Tool #14 (Extreme Principle): pin down how small the answer can possibly be, then show that small value is actually reachable. Two parts. First, find a hard floor: the two hundreds digits are different and each at least $1$, so the sum $S$ can't be too small. That floor lets Tool #3 (Eliminate Possibilities) knock out the digit sums $1$, $2$, and $3$. Second, Tool #6 (Guess and Check) builds one concrete pair of numbers whose sum has digit sum $4$, proving $4$ is attainable. A bound plus a matching example pins the minimum exactly.
Execute — Answer: B
4.NBT.A.2 Step 1 How small can S be?
- Each number is three digits, so each hundreds digit is at least $1$.
- The two hundreds digits must be different, so the smallest they can be is $1$ and $2$.
- That means the two hundreds digits add to at least $3$, so the two numbers add to at least $300$.
- Therefore $S$ is at least $300$.
💡 Two different non-zero hundreds digits are at smallest a $1$ and a $2$, which already drags the total up to the $300$s.
2.NBT.A.1 Step 2 Rule out digit sums 1, 2, 3
- Since $S \ge 300$ and $S$ has three digits, its hundreds digit is at least $3$.
- Among all three-digit numbers that are $300$ or more, the only one whose digits add to $3$ or less is $300$ itself.
- But $S = 300$ would need the tens column and the units column to each total $0$, which forces a tens digit of $0$ in both numbers (and likewise units) — repeated digits, which is not allowed.
- So $S$ cannot have a digit sum of $1$, $2$, or $3$.
💡 To make every digit of $S$ tiny you would need zeros stacked on zeros, but the all-different rule blocks reusing $0$ twice.
4.NBT.B.4 Step 3 Build a sum with digit sum 4
- Now show $4$ is reachable.
- Try $104 + 296$.
- The six digits used are $1, 0, 4$ and $2, 9, 6$ — all different.
- The sum is $104 + 296 = 400$, whose digits add to $4 + 0 + 0 = 4$.
- Since $4$ is reachable and we already ruled out $1$, $2$, and $3$, the smallest possible digit sum of $S$ is $4$.
- The answer is (B).
💡 Carries let the tens and units columns roll over to $0$, so only the hundreds digit is left to count.
4.NBT.A.2 Each number is three digits, so each hundreds digit is at least $1$. The two hun 2.NBT.A.1 Since $S \ge 300$ and $S$ has three digits, its hundreds digit is at least $3$. 4.NBT.B.4 Now show $4$ is reachable. Try $104 + 296$. The six digits used are $1, 0, 4$ an Review
Reasonableness: The example $104 + 296 = 400$ checks out by direct addition, its six digits $\{0,1,2,4,6,9\}$ are all different, and $400$ is a three-digit number with digit sum $4$. The bound says nothing below $4$ is possible, and the example hits exactly $4$, so the two halves agree. The losing choices fit the story: $(A)\,1$ is impossible because $S \ge 300$; $(D)\,15$ and $(E)\,20$ are far larger than needed.
Alternative: Use remainders mod $9$: a number and its digit sum leave the same remainder when divided by $9$. To get digit sum $4$ we need $S \equiv 4 \pmod 9$, and since $S$ equals the sum of the two addends, we need the six addend digits to total $4$ more than a multiple of $9$. The digits $\{0,1,2,4,6,9\}$ sum to $22 \equiv 4 \pmod 9$, matching $S = 400$. This confirms $4$ is consistent and reachable.
CCSS standards used (min grade 4)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Reasoning that two different non-zero hundreds digits are at least $1$ and $2$, so $S \ge 300$.)2.NBT.A.1Understand that the three digits of a three-digit number represent hundreds, tens, and ones (Reading each place of $S$ separately to show that digit sums $1$, $2$, $3$ would force repeated zeros.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding $104 + 296 = 400$ with carries to produce a sum whose digits total $4$.)
⭐ To make a sum's digits small, use carries to roll columns over to zero — but two different non-zero leading digits keep the sum stuck in the 300s, so 4 is as low as it goes.
⭐ To make a sum's digits small, use carries to roll columns over to zero — but two different non-zero leading digits keep the sum stuck in the 300s, so 4 is as low as it goes.
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