AMC 10 · 2021 · #3
Grade 6 number-theoryProblem
The sum of two natural numbers is 17,402. One of the two numbers is divisible by 10. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
Pick an answer.
AMC 10 2021 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 natural numbers add to $17{,}402$. The bigger one ends in $0$ (it is a multiple of $10$). Erasing that bigger number's units digit leaves the smaller number. Find the difference between the two numbers.
Givens: Bigger $+$ smaller $= 17{,}402$; Bigger is a multiple of $10$ (its ones digit is $0$); Erasing the bigger number's ones digit gives the smaller number — so bigger $= 10 \times$ smaller; Answer choices: (A) $10{,}272$, (B) $11{,}700$, (C) $13{,}362$, (D) $14{,}238$, (E) $15{,}426$
Unknowns: Bigger $-$ smaller
Understand
Restated: Two natural numbers add to $17{,}402$. The bigger one ends in $0$ (it is a multiple of $10$). Erasing that bigger number's units digit leaves the smaller number. Find the difference between the two numbers.
Givens: Bigger $+$ smaller $= 17{,}402$; Bigger is a multiple of $10$ (its ones digit is $0$); Erasing the bigger number's ones digit gives the smaller number — so bigger $= 10 \times$ smaller; Answer choices: (A) $10{,}272$, (B) $11{,}700$, (C) $13{,}362$, (D) $14{,}238$, (E) $15{,}426$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities, #6 Guess and Check
Tool #7 (Subproblems) is the spine: name the smaller number $s$; then the bigger is $10s$ (since erasing the ones digit is the same as dividing by $10$). The sum condition becomes one tiny equation $11s = 17{,}402$, and the difference is $9s$. Tool #3 (Eliminate) acts as a fast safety net — the difference ends in $9s$, whose ones digit must be $8$ (since $s$ ends in $2$). Only choice (D) $14{,}238$ ends in $8$.
Execute — Answer: D
5.NBT.A.1 Step 1 - Let $s$ be the smaller number.
- Because erasing the bigger number's ones digit ($0$) gives $s$, the bigger number is exactly $10s$ (a digit shifted left by one place is multiplication by $10$).
💡 Sliding all digits one place to the left multiplies the number by $10$ — Grade 5 "a digit's place is $10$ times the place to its right".
6.EE.B.7 Step 2 - Use the sum condition.
- $\text{bigger} + \text{smaller} = 10s + s = 11s$, and that equals $17{,}402$.
- So $s = 17{,}402 \div 11 = 1{,}582$.
💡 One unknown, one tidy equation $11s = 17{,}402$ — Grade 6 "solve $px = q$" by dividing.
4.NBT.B.4 Step 3 - Compute the two numbers.
- Smaller $= 1{,}582$, bigger $= 10 \times 1{,}582 = 15{,}820$.
- The difference is $15{,}820 - 1{,}582 = 14{,}238$, which is choice (D).
💡 Subtracting two multi-digit whole numbers with regrouping — Grade 4 standard algorithm.
4.NBT.A.2 Step 4 - Quick safety check by ones digit.
- Smaller ends in $2$, bigger ends in $0$.
- Ones digit of (bigger $-$ smaller) is $0 - 2 = 8$ (borrowing).
- Only (D) $14{,}238$ ends in $8$, confirming the answer without redoing the subtraction.
💡 Comparing ones digits is fast — Grade 4 "compare multi-digit whole numbers using digit positions".
5.NBT.A.1 Let $s$ be the smaller number. Because erasing the bigger number's ones digit ($ 6.EE.B.7 Use the sum condition. $\text{bigger} + \text{smaller} = 10s + s = 11s$, and tha 4.NBT.B.4 Compute the two numbers. Smaller $= 1{,}582$, bigger $= 10 \times 1{,}582 = 15{, 4.NBT.A.2 Quick safety check by ones digit. Smaller ends in $2$, bigger ends in $0$. Ones Review
Reasonableness: Verify the trio: smaller $+$ bigger $= 1{,}582 + 15{,}820 = 17{,}402 \checkmark$. Erasing the ones digit of $15{,}820$ gives $1{,}582 \checkmark$. Difference $14{,}238$ is roughly $9/11$ of $17{,}402 \approx 14{,}238 \checkmark$. The number lines up with choice (D), and no other choice has the right ones digit.
Alternative: Tool #6 (Guess and Check) on round trial values. Guess smaller $= 1{,}500$: bigger $= 15{,}000$, sum $= 16{,}500$, too small. Guess smaller $= 1{,}600$: bigger $= 16{,}000$, sum $= 17{,}600$, too big. So smaller is between, closer to $1{,}600$. Try $1{,}582$: sum $= 17{,}402 \checkmark$. Same answer.
CCSS standards used (min grade 6)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Reading off the ones digits of bigger and smaller to compute the ones digit of the difference.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Computing $15{,}820 - 1{,}582 = 14{,}238$ by the standard algorithm.)5.NBT.A.1Recognize that a digit in one place represents ten times as much as to its right (Justifying that bigger $= 10 \times$ smaller because the digits slide one place to the left.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Setting up $11s = 17{,}402$ from the sum condition and dividing to find $s = 1{,}582$.)
⭐ This AMC 10 problem only needs Grade 6 "sliding digits left is multiplying by ten" — once you call the smaller number $s$, the bigger is $10s$, and $11s = 17{,}402$ gives $s = 1{,}582$, so the difference is $14{,}238$.
⭐ This AMC 10 problem only needs Grade 6 "sliding digits left is multiplying by ten" — once you call the smaller number $s$, the bigger is $10s$, and $11s = 17{,}402$ gives $s = 1{,}582$, so the difference is $14{,}238$.
More like this
Same archetype — closest grade level first.