AMC 10 · 2021 · #3

Grade 6 number-theory

Archetype Arithmetic Expression Decomposition

place-valuedigit-decompositionlinear-equations-one-varmulti-digit-arithmetic convert-to-algebraidentify-subproblems ↑ Prerequisites: place-value

📏 Short solution 💡 2 insights

Problem

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.

(A)

$~10{,}272$

(B)

$~11{,}700$

(C)

$~13{,}362$

(D)

$~14{,}238$

(E)

$~15{,}426$

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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

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

#7 Identify Subproblems 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 $10$).

$$\text{smaller} = s, \quad \text{bigger} = 10s$$

💡 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".

#7 Identify Subproblems 6.EE.B.7 Step 2

Use the sum condition.
$\text{bigger} + \text{smaller} = 10s + s = 11s$, and that equals $17{,}402$.
So $s = 17{,}402 \div 11 = 1{,}582$.

$$11s = 17{,}402 \;\Rightarrow\; s = \dfrac{17{,}402}{11} = 1{,}582$$

💡 One unknown, one tidy equation $11s = 17{,}402$ — Grade 6 "solve $px = q$" by dividing.

#7 Identify Subproblems 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).

$$15{,}820 - 1{,}582 = 14{,}238 \;\Rightarrow\; \textbf{(D)}$$

💡 Subtracting two multi-digit whole numbers with regrouping — Grade 4 standard algorithm.

#3 Eliminate Possibilities 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.

$$\text{ones digit of difference} = 8 \;\Rightarrow\; \textbf{(D)}$$

💡 Comparing ones digits is fast — Grade 4 "compare multi-digit whole numbers using digit positions".

[1] #7 5.NBT.A.1 Let $s$ be the smaller number. Because erasing the bigger number's ones digit ($

[2] #7 6.EE.B.7 Use the sum condition. $\text{bigger} + \text{smaller} = 10s + s = 11s$, and tha

[3] #7 4.NBT.B.4 Compute the two numbers. Smaller $= 1{,}582$, bigger $= 10 \times 1{,}582 = 15{,

[4] #3 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.2 Read 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.4 Fluently add and subtract multi-digit whole numbers (Computing $15{,}820 - 1{,}582 = 14{,}238$ by the standard algorithm.)
5.NBT.A.1 Recognize 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.7 Solve 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 $14{,}238$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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