AMC 10 · 2023 · #12

Grade 4 number-theory

divisibility-rulesdigit-decompositionsystematic-enumerationmodular-arithmetic caseworksystematic-enumerationidentify-subproblems ↑ Prerequisites: divisibility-rulesdigit-decomposition

📏 Long solution 💡 3 insights

📘 View easy version →

Problem

How many three-digit positive integers $N$ satisfy the following properties?

The number $N$ is divisible by $7$ .
The number formed by reversing the digits of $N$ is divisible by $5$ .

$\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Count the three-digit integers $N$ that satisfy two conditions simultaneously: $N$ itself is a multiple of $7$, and when you reverse $N$'s digits the resulting number is a multiple of $5$.

Givens: $N$ is a three-digit positive integer, so $100 \le N \le 999$; $N$ is divisible by $7$; If $N = \overline{abc}$ (digits $a, b, c$), then its reversal $\overline{cba}$ is divisible by $5$; Answer choices: (A) 13, (B) 14, (C) 15, (D) 16, (E) 17

Unknowns: The count of three-digit $N$ satisfying both conditions

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #2 Systematic List, #3 Eliminate Possibilities

Two conditions act on different digits, so Tool #7 (Identify Subproblems) splits the work: first use the reversal-divisibility-by-$5$ rule to pin down the hundreds digit of $N$, then use divisibility by $7$ to count how many of the resulting three-digit numbers survive. The divisibility-by-$5$ subproblem collapses to a single hundreds digit because $a \ne 0$. Tool #2 (Systematic List) underpins step two — listing multiples of $7$ in the locked-down range, then using Tool #3 (Eliminate) to cross-check the final count against the choices.

Execute — Answer: B

#7 Identify Subproblems 4.OA.B.4 Step 1

Subproblem A: what does "reversed number divisible by $5$" force?
Write $N$ as $\overline{abc}$.
Reversed, the number is $\overline{cba}$, whose units digit is $a$.
A number is divisible by $5$ only when its units digit is $0$ or $5$.

$$a \in \{0, 5\}$$

💡 The divisibility-by-$5$ rule (units digit $0$ or $5$) is the very first divisibility rule kids meet — pure Grade 4 number-sense.

#7 Identify Subproblems 2.NBT.A.1 Step 2

Now apply the three-digit constraint.
The hundreds digit of a three-digit number can never be $0$, so the leading $a$ must be $5$.
Every valid $N$ starts with a $5$.

$$a = 5 \;\Rightarrow\; 500 \le N \le 599$$

💡 The Grade 2 place-value rule "hundreds digit can't be 0" knocks $a = 0$ out and pins the whole search to the $500$s.

#7 Identify Subproblems 4.NBT.B.6 Step 3

Subproblem B: how many multiples of $7$ lie in $[500, 599]$?
Find the first multiple of $7$ at or above $500$ by dividing: $500 \div 7 = 71\text{ r } 3$, so the next multiple is $7 \times 72 = 504$.

$$\lceil 500/7 \rceil = 72, \quad 7 \times 72 = 504$$

💡 Dividing with remainder (Grade 4) tells you exactly where the next clean multiple lands — no guess-and-check needed.

#7 Identify Subproblems 4.NBT.B.6 Step 4

Find the last multiple of $7$ at or below $599$ the same way: $599 \div 7 = 85\text{ r } 4$, so the largest multiple is $7 \times 85 = 595$.

$$\lfloor 599/7 \rfloor = 85, \quad 7 \times 85 = 595$$

💡 Same divide-with-remainder trick from the upper end — the last clean multiple is one step before $599$.

#2 Systematic List 4.OA.C.5 Step 5

List the surviving numbers in order — Tool #2.
They are $7 \times 72, 7 \times 73, \dots, 7 \times 85$: an arithmetic list indexed by the integers $72, 73, \dots, 85$.

$$504, 511, 518, 525, 532, 539, 546, 553, 560, 567, 574, 581, 588, 595$$

💡 Following the rule "add $7$ each time, stop at $595$" generates the list in order — Grade 4 pattern generation.

#2 Systematic List 3.OA.D.8 Step 6

Count the list.
The number of integers from $72$ to $85$ inclusive is $85 - 72 + 1 = 14$.

$$85 - 72 + 1 = 14$$

💡 "Last minus first plus one" is the basic Grade 3 counting move for consecutive integers.

#3 Eliminate Possibilities 4.OA.B.4 Step 7

Cross-check the candidate $N = 504$.
Its reversal is $405$, units digit $5$, divisible by $5$ — good.
And $504 = 7 \times 72$ is divisible by $7$.
Both conditions hold; the count of $14$ matches answer (B).

$$504 \to 405 = 5 \times 81; \quad 504 = 7 \times 72 \;\Rightarrow\; \textbf{(B)}$$

💡 Verifying one item from the list against both conditions confirms the whole list is on the right track.

[1] #7 4.OA.B.4 Subproblem A: what does "reversed number divisible by $5$" force? Write $N$ as $

[2] #7 2.NBT.A.1 Now apply the three-digit constraint. The hundreds digit of a three-digit number

[3] #7 4.NBT.B.6 Subproblem B: how many multiples of $7$ lie in $[500, 599]$? Find the first mult

[4] #7 4.NBT.B.6 Find the last multiple of $7$ at or below $599$ the same way: $599 \div 7 = 85\t

[5] #2 4.OA.C.5 List the surviving numbers in order — Tool #2. They are $7 \times 72, 7 \times 7

[6] #2 3.OA.D.8 Count the list. The number of integers from $72$ to $85$ inclusive is $85 - 72 +

[7] #3 4.OA.B.4 Cross-check the candidate $N = 504$. Its reversal is $405$, units digit $5$, div

Review

Reasonableness: Sanity check on the count: between $500$ and $599$ the multiples of $7$ should number roughly $\tfrac{100}{7} \approx 14.3$, so $14$ is the right magnitude. End-point check: $504$ reverses to $405 = 5 \times 81$ and $595$ reverses to $595$ itself (palindrome multiple of $5 \cdot 7 = 35$, since $595 = 17 \times 35$), both pass. The first and last entries of the list are real, no boundary error.

Alternative: Tool #2 (Systematic List) directly: list every multiple of $7$ from $100$ to $999$ (that's $128$ numbers), then keep only those whose hundreds digit is $5$. The hundreds-digit-$5$ slice of that list is exactly $504$ to $595$ stepping by $7$ — $14$ entries. Same answer, more work, but no divisibility rule needed.

CCSS standards used (min grade 4)

2.NBT.A.1 Understand that the three digits of a three-digit number represent hundreds, tens, and ones (Knowing the hundreds digit of a three-digit number must be at least $1$, which forces $a = 5$ (not $0$).)
3.OA.D.8 Solve two-step word problems using four operations within 100 (Counting consecutive integers $72$ through $85$ as $85 - 72 + 1 = 14$.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Applying the divisibility-by-$5$ rule (units digit $0$ or $5$) to the reversed number, and verifying that $504$ is a multiple of both $7$ and that its reversal is a multiple of $5$.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Listing the multiples of $7$ in the $500$s by stepping in arithmetic progression: $504, 511, 518, \dots, 595$.)
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends (Computing $500 \div 7 = 71 \text{ r } 3$ and $599 \div 7 = 85 \text{ r } 4$ to bracket the multiples of $7$ in the range.)

⭐ Two divisibility conditions split into two small steps: the "reversed is a multiple of $5$" rule pins down the leading digit (it has to be $5$), and then you just count the multiples of $7$ between $500$ and $599$ — fourteen of them. Grade 4 divisibility rules carry the whole problem.