AMC 10 · 2016 · #6
Easy mode Grade 4Ximena writes the whole numbers from 1 to 30. Emilio copies her list, but he changes every digit 2 into a digit 1. For example, 2 becomes 1, 12 becomes 11, and 20 becomes 10. Ximena adds up all of her numbers, and Emilio adds up all of his. How much bigger is Ximena's total than Emilio's?
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: Ximena writes the whole numbers $1$ through $30$. Emilio copies them but turns every digit $2$ into a digit $1$. Find how much bigger Ximena's total is than Emilio's total.
Givens: Ximena lists the whole numbers $1, 2, 3, \ldots, 30$; Emilio copies the same list but replaces each digit $2$ with the digit $1$; Ximena adds her numbers; Emilio adds his numbers
Unknowns: By how much Ximena's sum exceeds Emilio's sum
Understand
Restated: Ximena writes the whole numbers $1$ through $30$. Emilio copies them but turns every digit $2$ into a digit $1$. Find how much bigger Ximena's total is than Emilio's total.
Givens: Ximena lists the whole numbers $1, 2, 3, \ldots, 30$; Emilio copies the same list but replaces each digit $2$ with the digit $1$; Ximena adds her numbers; Emilio adds his numbers
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
Adding all thirty numbers twice is slow and easy to slip on. Tool #16 (Change Focus) drops both full sums and looks only at the gap between them: a number changes its value only where a digit $2$ becomes a digit $1$, so the whole difference is built from those single-digit drops. Tool #7 (Identify Subproblems) then splits those drops by place value — tens-place $2$s cost $10$ each, ones-place $2$s cost $1$ each — because the two cases shrink the number by different amounts. Tool #2 (Make a Systematic List) just counts how many $2$s sit in each place across $1$ to $30$.
Execute — Answer: D
4.OA.A.3 Step 1 Track only the drops
- Every number Emilio writes is either identical to Ximena's or smaller, and it only changes where a digit $2$ turned into a $1$.
- So the difference between the two sums equals the total amount lost across all the $2 \to 1$ swaps.
- Ignore the rest of the digits entirely.
💡 If two lists differ only in a few spots, the gap in their sums is just the change at those spots.
1.NBT.B.2 Step 2 Split by place value
- A digit $2$ is worth different amounts depending on its place.
- A $2$ in the tens place stands for $20$; turning it into a $1$ makes it $10$, a drop of $10$.
- A $2$ in the ones place stands for $2$; turning it into a $1$ makes it $1$, a drop of $1$.
- So handle tens-place $2$s and ones-place $2$s separately.
💡 The same digit $2$ is worth ten times more in the tens place than in the ones place.
3.OA.A.1 Step 3 Count the tens-place 2s
- List the numbers from $1$ to $30$ whose tens digit is $2$: these are $20, 21, 22, 23, 24, 25, 26, 27, 28, 29$ — exactly the ten numbers in the twenties.
- Each one loses $10$, so together they lose $10 \times 10 = 100$.
💡 Every number in the twenties has a tens-place $2$, and there are ten of them.
3.OA.A.1 Step 4 Count the ones-place 2s
- List the numbers from $1$ to $30$ whose ones digit is $2$: these are $2, 12, 22$ — three of them.
- Each loses $1$, so together they lose $3 \times 1 = 3$.
- (The number $22$ already had its tens $2$ counted in the previous step; here we count its ones $2$ separately.)
💡 Only $2$, $12$, and $22$ end in a $2$, and each loses just one.
4.NBT.B.4 Step 5 Add the two drops
- The total difference is the tens-place loss plus the ones-place loss: $100 + 3 = 103$.
- Ximena's sum is $103$ larger than Emilio's, which is choice $\textbf{(D)}$.
💡 Add the big losses and the small losses to get the whole gap.
4.OA.A.3 Every number Emilio writes is either identical to Ximena's or smaller, and it on 1.NBT.B.2 A digit $2$ is worth different amounts depending on its place. A $2$ in the tens 3.OA.A.1 List the numbers from $1$ to $30$ whose tens digit is $2$: these are $20, 21, 22 3.OA.A.1 List the numbers from $1$ to $30$ whose ones digit is $2$: these are $2, 12, 22$ 4.NBT.B.4 The total difference is the tens-place loss plus the ones-place loss: $100 + 3 = Review
Reasonableness: The tens-place loss must be the dominant piece, and $100$ from ten numbers each dropping $10$ is exactly that, so the answer should be just above $100$ — and $103$ is. Spot-check the ones: $22$ becomes $11$, a drop of $11 = 10 + 1$, which correctly counts once in the tens total and once in the ones total. Choices $13$ and $26$ are far too small (they ignore the ten numbers in the twenties), and $110$ would wrongly assume ten ones-place $2$s, so $103$ is the only sensible value.
Alternative: Compute both sums directly. Ximena's sum is $1 + 2 + \cdots + 30 = \frac{30 \cdot 31}{2} = 465$. For Emilio, rewrite the twenties $20\text{–}29$ as $10\text{–}19$ (lowering the tens digit) and rewrite $2 \to 1$ and $12 \to 11$; his sum works out to $362$. Then $465 - 362 = 103$, the same answer (D).
CCSS standards used (min grade 4)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Reframing the problem as the total value lost across all digit changes instead of computing two full sums.)1.NBT.B.2Understand that the two digits of a two-digit number represent tens and ones (Recognizing that a $2$ in the tens place drops by $10$ while a $2$ in the ones place drops by $1$.)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Multiplying the count of tens-place $2$s by $10$ and the count of ones-place $2$s by $1$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the tens-place loss $100$ and the ones-place loss $3$ to get $103$.)
⭐ Only the digit $2$s change, so just add up what each one loses — $10$ for every tens-place $2$ and $1$ for every ones-place $2$.
⭐ Only the digit $2$s change, so just add up what each one loses — $10$ for every tens-place $2$ and $1$ for every ones-place $2$.
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