AMC 8 · 2025 · #23
Grade 5 number-theoryProblem
How many four-digit numbers have all three of the following properties?
(I) The tens and ones digit are both 9.
(II) The number is 1 less than a perfect square.
(III) The number is the product of exactly two prime numbers.
Pick an answer.
AMC 8 2025 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 four-digit whole numbers $N$ (from $1000$ to $9999$) that satisfy all three of: (I) the last two digits of $N$ are both $9$, (II) $N + 1$ is a perfect square, and (III) $N$ has exactly two prime factors when written as a product of primes.
Givens: $N$ is a four-digit number, so $1000 \le N \le 9999$; The tens digit and the ones digit of $N$ are both $9$ (so $N$ ends in $99$); $N = k^{2} - 1$ for some positive integer $k$; $N = p \times q$ where $p$ and $q$ are prime numbers (counted with multiplicity, $N$ has exactly two prime factors); Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: The number of four-digit integers $N$ that satisfy all three conditions
Understand
Restated: Count the four-digit whole numbers $N$ (from $1000$ to $9999$) that satisfy all three of: (I) the last two digits of $N$ are both $9$, (II) $N + 1$ is a perfect square, and (III) $N$ has exactly two prime factors when written as a product of primes.
Givens: $N$ is a four-digit number, so $1000 \le N \le 9999$; The tens digit and the ones digit of $N$ are both $9$ (so $N$ ends in $99$); $N = k^{2} - 1$ for some positive integer $k$; $N = p \times q$ where $p$ and $q$ are prime numbers (counted with multiplicity, $N$ has exactly two prime factors); Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
Combining conditions (I) and (II) is a small subproblem: $N$ ends in $99$ means $N + 1 = k^{2}$ ends in $00$, which forces $k$ to be a multiple of $10$. That collapses the search from thousands of four-digit numbers down to a tiny finite list of $k$ values (Tool #7 splits the work cleanly). Tool #2 (Systematic List) then enumerates every candidate $k = 40, 50, \ldots, 100$ in order. For each, condition (III) is checked by noting $k^{2}-1 = (k-1)(k+1)$ — both factors must be prime — which is Tool #3 (Eliminate Possibilities) crossing off any $k$ where $k-1$ or $k+1$ is composite. No algebra needed beyond grade-school multiplication and primality checks.
Execute — Answer: B
5.NBT.A.2 Step 1 - Combine conditions (I) and (II).
- "Ends in $99$" means $N + 1$ ends in $00$, i.e., $N + 1$ is a multiple of $100$.
- Since $N + 1 = k^{2}$, the perfect square $k^{2}$ must end in $00$.
- The only way a square ends in $00$ is when the integer being squared ends in $0$, so $k$ must be a multiple of $10$.
💡 Patterns in trailing zeros of place value are a Grade 5 idea: a square ending in $00$ has to come from a number ending in $0$.
5.NBT.B.5 Step 2 - Bound $k$ using the four-digit window.
- We need $1000 \le N \le 9999$, which gives $1001 \le k^{2} \le 10000$.
- Combined with $k = 10m$, the only candidates are $k = 40, 50, 60, 70, 80, 90, 100$ (since $30^{2}=900$ is too small and $110^{2}=12100$ is too big).
💡 Multi-digit multiplication ($40^{2}, 50^{2}, \ldots$) is a Grade 5 fluency standard, plenty to range-check the candidates.
4.OA.B.4 Step 3 - Use the difference-of-squares factoring $k^{2} - 1 = (k-1)(k+1)$.
- This is the key observation: $N$ is already the product of $k-1$ and $k+1$, so $N$ is a product of exactly two primes precisely when both $k-1$ and $k+1$ are themselves prime.
- Any composite factor would introduce extra primes.
💡 Verifying $(k-1)(k+1) = k^{2}-1$ is just multiplication, and "is it prime?" is the Grade 4 prime/composite skill.
4.OA.B.4 Step 4 - Now run through the seven candidates and eliminate each one whose $k-1$ or $k+1$ is composite.
- Only $k = 60$ survives because $59$ and $61$ are both prime (twin primes).
- The corresponding $N = 59 \times 61 = 3599$.
💡 Testing each $k-1$ and $k+1$ for primality by trial division is exactly the Grade 4 "determine prime or composite" task.
4.OA.B.4 Step 5 - Exactly one four-digit number, $N = 3599 = 59 \times 61$, satisfies all three conditions.
- The count is $1$, which is choice (B).
💡 Tallying the surviving candidates from the systematic list is Grade 4 counting.
5.NBT.A.2 Combine conditions (I) and (II). "Ends in $99$" means $N + 1$ ends in $00$, i.e. 5.NBT.B.5 Bound $k$ using the four-digit window. We need $1000 \le N \le 9999$, which give 4.OA.B.4 Use the difference-of-squares factoring $k^{2} - 1 = (k-1)(k+1)$. This is the ke 4.OA.B.4 Now run through the seven candidates and eliminate each one whose $k-1$ or $k+1$ 4.OA.B.4 Exactly one four-digit number, $N = 3599 = 59 \times 61$, satisfies all three co Review
Reasonableness: Sanity check the winner: $3599$ ends in $99$ (condition I), $3599 + 1 = 3600 = 60^{2}$ is a perfect square (condition II), and $3599 = 59 \times 61$ with both factors prime, so exactly two prime factors (condition III). All three hold. Also, the answer of $1$ is consistent with how AMC 8 typically rewards twin-prime-style observations — twin primes thin out fast, so finding only one match in a small window is expected, not suspicious.
Alternative: Tool #5 (Look for a Pattern): instead of factoring $k^{2}-1$, you could compute each candidate $N$ directly ($1599, 2499, 3599, 4899, 6399, 8099, 9999$) and try to factor each by trial division up to $\sqrt{N} \approx 100$. This is slower and more arithmetic-heavy, but a student who hasn't yet noticed the $(k-1)(k+1)$ trick can still finish the problem with patience and a primality check.
CCSS standards used (min grade 5)
5.NBT.A.2Explain patterns in number of zeros and placement of decimal point (Recognizing that a perfect square ending in $00$ must come from an integer ending in $0$, so $k$ is a multiple of $10$.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Computing the squares $40^{2}, 50^{2}, \ldots, 100^{2}$ to confirm which values of $k$ keep $N = k^{2} - 1$ inside the four-digit window.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Checking for each candidate $k$ whether $k-1$ and $k+1$ are prime (twin primes), since $N = (k-1)(k+1)$ is the product of exactly two primes iff both factors are prime.)
⭐ This AMC 8 problem only needs Grade 5 place-value patterns plus the Grade 4 "is this number prime?" check you already know!
⭐ This AMC 8 problem only needs Grade 5 place-value patterns plus the Grade 4 "is this number prime?" check you already know!
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