AMC 10 · 2023 · #12
Grade 4 number-theoryProblem
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.
Pick an answer.
AMC 10 2023 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: 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
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
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
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 $5$.
💡 The divisibility-by-$5$ rule (units digit $0$ or $5$) is the very first divisibility rule kids meet — pure Grade 4 number-sense.
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 $5$.
💡 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.
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 $7 \times 72 = 504$.
💡 Dividing with remainder (Grade 4) tells you exactly where the next clean multiple lands — no guess-and-check needed.
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 $7 \times 85 = 595$.
💡 Same divide-with-remainder trick from the upper end — the last clean multiple is one step before $599$.
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 $72, 73, \dots, 85$.
💡 Following the rule "add $7$ each time, stop at $595$" generates the list in order — Grade 4 pattern generation.
3.OA.D.8 Step 6 - Count the list.
- The number of integers from $72$ to $85$ inclusive is $85 - 72 + 1 = 14$.
💡 "Last minus first plus one" is the basic Grade 3 counting move for consecutive integers.
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).
💡 Verifying one item from the list against both conditions confirms the whole list is on the right track.
4.OA.B.4 Subproblem A: what does "reversed number divisible by $5$" force? Write $N$ as $ 2.NBT.A.1 Now apply the three-digit constraint. The hundreds digit of a three-digit number 4.NBT.B.6 Subproblem B: how many multiples of $7$ lie in $[500, 599]$? Find the first mult 4.NBT.B.6 Find the last multiple of $7$ at or below $599$ the same way: $599 \div 7 = 85\t 4.OA.C.5 List the surviving numbers in order — Tool #2. They are $7 \times 72, 7 \times 7 3.OA.D.8 Count the list. The number of integers from $72$ to $85$ inclusive is $85 - 72 + 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.1Understand 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.8Solve two-step word problems using four operations within 100 (Counting consecutive integers $72$ through $85$ as $85 - 72 + 1 = 14$.)4.OA.B.4Find 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.5Generate 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.6Find 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.
⭐ 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.