AMC 8 · 2013 · #13
Easy mode Grade 5Problem
Clara was adding up a list of scores. Each score is a two-digit number, like 52 or 74.
When she wrote down one of those scores, she swapped its tens digit and its ones digit by accident. For example, 52 might have become 25, or 74 might have become 47.
So her total came out wrong. Compared to the correct total, her wrong total is off by some amount (it could be too big or too small).
Which of the following numbers could be the size of that difference?
Pick an answer.
AMC 8 2013 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: Clara added up several test scores but, on one score, she accidentally swapped the tens digit and the units digit (for example, she wrote $72$ as $27$). Because of that single swap, her total came out different from the true total. Which of the five answer choices could that difference be?
Givens: Exactly one two-digit score had its tens and units digits reversed; All other scores were added correctly; Answer choices: (A) $45$, (B) $46$, (C) $47$, (D) $48$, (E) $49$
Unknowns: Which of the five differences could actually occur from a single digit swap
Understand
Restated: Clara added up several test scores but, on one score, she accidentally swapped the tens digit and the units digit (for example, she wrote $72$ as $27$). Because of that single swap, her total came out different from the true total. Which of the five answer choices could that difference be?
Givens: Exactly one two-digit score had its tens and units digits reversed; All other scores were added correctly; Answer choices: (A) $45$, (B) $46$, (C) $47$, (D) $48$, (E) $49$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
Instead of jumping into algebra, try small concrete cases first (Tool #9). Reverse the digits of a few two-digit numbers — $72 \to 27$, $63 \to 36$, $41 \to 14$ — and write down each difference. Tool #5 (Look for a Pattern) spots that every difference is a multiple of $9$. Tool #3 (Eliminate Possibilities) then sweeps the five answer choices and keeps only the multiple of $9$. This path uses Grade 4 multiples / place-value reasoning instead of Grade 6 algebra.
Execute — Answer: A
4.NBT.B.4 Step 1 - Try a few easy examples by hand.
- Pick any two-digit score, reverse its digits, and find the difference.
- Doing this for three numbers gives enough data to spot a pattern.
💡 Working a smaller, concrete version is the Grade 4 "add and subtract multi-digit whole numbers" skill — no algebra needed yet.
4.OA.B.4 Step 2 - Look at the three differences: $45,\ 27,\ 27$.
- Each one is a multiple of $9$: $45 = 9 \times 5$ and $27 = 9 \times 3$.
- Try one more case to be sure — say $85 - 58 = 27 = 9 \times 3$.
- The pattern holds: a digit-swap difference is always a multiple of $9$.
💡 Recognizing that every result is a multiple of $9$ is the Grade 4 "factors and multiples" idea applied to a pattern of numbers.
5.NBT.A.1 Step 3 - See *why* the pattern works using place value.
- A two-digit number with tens digit $t$ and units digit $u$ has value $10t + u$.
- Swapping the digits gives $10u + t$.
- Subtracting: $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$ — a multiple of $9$, no matter which digits Clara used.
💡 Reading a two-digit number as $10t + u$ is the Grade 5 place-value rule "each digit is $10$ times the place to its right."
4.OA.B.4 Step 4 - Apply Tool #3 (Eliminate Possibilities) to the five choices: keep only multiples of $9$.
- Quick check via the divisibility-by-$9$ rule (digits sum to a multiple of $9$): $4+5=9$ ✓, $4+6=10$, $4+7=11$, $4+8=12$, $4+9=13$.
- Only $45$ survives.
💡 Checking each choice against the divisibility rule for $9$ is exactly the Grade 4 "multiples" skill in action.
4.OA.B.4 Step 5 - Confirm that the surviving value $45$ really *can* happen.
- From step 3, the difference is $9(t-u)$, so $9(t-u) = 45$ means $|t - u| = 5$.
- Plenty of digit pairs work (e.g., $7$ and $2$): $72 - 27 = 45$.
- So $45$ is achievable.
💡 Producing a concrete example confirms the answer survives both the pattern check and a real-world check.
4.NBT.B.4 Try a few easy examples by hand. Pick any two-digit score, reverse its digits, a 4.OA.B.4 Look at the three differences: $45,\ 27,\ 27$. Each one is a multiple of $9$: $4 5.NBT.A.1 See *why* the pattern works using place value. A two-digit number with tens digi 4.OA.B.4 Apply Tool #3 (Eliminate Possibilities) to the five choices: keep only multiples 4.OA.B.4 Confirm that the surviving value $45$ really *can* happen. From step 3, the diff Review
Reasonableness: Every test case we tried — $72 \to 27$, $63 \to 36$, $41 \to 14$, $85 \to 58$ — gave a difference that is a multiple of $9$. The algebraic shortcut $9(t-u)$ explains why this *must* be true. Among the five choices, only $45$ is a multiple of $9$, and we exhibited a real swap ($72 \leftrightarrow 27$) that produces exactly $45$. The reasoning and the example both point to (A).
Alternative: Tool #6 (Guess and Check) on the answer choices: for each option, ask "can I find two digits $t, u$ with $10t + u - (10u + t) = $ that option?" The expression simplifies to $9(t-u)$, so the question becomes "is the option divisible by $9$?" Only $45$ qualifies, matching the pattern-based path.
CCSS standards used (min grade 5)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Computing the small concrete cases ($72 - 27 = 45$, $63 - 36 = 27$, $41 - 14 = 27$) that seed the pattern search.)4.OA.B.4Find all factor pairs for a whole number; recognize that a whole number is a multiple of each of its factors (Recognizing each difference as a multiple of $9$, and screening the five answer choices via the divisibility-by-$9$ rule.)5.NBT.A.1Recognize that in a multi-digit number, a digit in one place represents $10$ times as much as it represents in the place to its right (Writing a two-digit number as $10t + u$ and the reversed number as $10u + t$ to explain why the difference is always $9(t-u)$.)
⭐ Swapping the two digits of a number always changes it by a multiple of $9$ — a Grade 5 place-value fact, not an AMC mystery!
⭐ Swapping the two digits of a number always changes it by a multiple of $9$ — a Grade 5 place-value fact, not an AMC mystery!
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