AMC 8 · 2025 · #23
Easy mode Grade 5Problem
We are looking for four-digit numbers that have all three of the following properties.
(I) When you look at the number, the last two digits (the tens digit and the ones digit) are both 9. So the number looks like __99.
(II) If you add 1 to the number, the result is a perfect square (a number like 1,4,9,16,25,… that you get by multiplying an integer by itself).
(III) The number can be written as exactly two prime numbers multiplied together. (A prime number is a whole number bigger than 1 whose only divisors are 1 and itself, like 2,3,5,7,11,….)
How many four-digit numbers have all three of these properties?
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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