2025 AMC 8 Problem 23

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To help you master 2025 AMC 8 Problem 23, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

Want more practice? Scroll down to download Think Academy’s AMC 8 practice packs and keep building your AMC 8 skills.

2025 AMC8 Problem 23

Solution

Solution 1

Factorization and Case Checking Method

Express the number as $N=k^2-1$ and factor it as $(k-1)(k+1)$. For $N$ to have exactly two prime factors, both $k-1$ and $k+1$ must be prime. The condition that $N$ ends in 99 will limit $k$ to a short list of multiples of 10.

Difference of Squares Formula: $a^2-b^2=(a-b)(a+b)$. This fundamental identity allows us to transform ‘one less than a perfect square’ into a product of two factors.

Step 1: Set up the algebraic equation for the number

We are given that the number is 1 less than a perfect square. Let the four-digit number be $N$ and the perfect square be $k^2$. This gives us a direct equation for $N$.

$$N=k^2-1$$

Step 2: Use the last digits to restrict the perfect square

The problem states that the tens and ones digits of $N$ are both 9, meaning $N$ ends in 99. Therefore, $N+1$ must end in 00. Since $N+1=k^2$, the perfect square $k^2$ is a multiple of 100. This implies its square root $k$ must be a multiple of 10.

$$N\equiv 99(\mathrm{mod}100)$$

$$N+1\equiv 0(\mathrm{mod}100)$$

$$k^2\equiv 0(\mathrm{mod}100)$$

$$k\equiv 0(\mathrm{mod}10)$$

Step 3: Determine the possible range of values for $k$

Because $N$ is a four-digit number, it must be between 1000 and 9999. We can plug this range into our equation to find the bounds for $k^2$. Taking the square root gives us the bounds for $k$, allowing us to list all valid multiples of 10.

$$1000\le k^2-1\le 9999$$

$$1001\le k^2\le 10000$$

$$31.6<k\le 100$$

$$k\in \{40,50,60,70,80,90,100\}$$

Step 4: Use the prime factorization property

We are told $N$ is the product of exactly two prime numbers, $p$ and $q$. We can also factor $k^2-1$ using the difference of squares into $(k-1)(k+1)$. Because $k\ge 40$, neither factor is 1. The only way to split a product of two primes into two integer factors (with neither being 1) is if the factors are exactly the two primes themselves. Thus, $k-1$ and $k+1$ must both be prime numbers.

$$N=(k-1)(k+1)=p\cdot q$$

Step 5: Test the candidates for $k$

We test each candidate to see if both $k-1$ and $k+1$ are prime numbers. We can stop checking a pair as soon as we find one composite number.

$$k=40⟹39,41(39=3\times 13)$$

$$k=50⟹49,51(49=7\times 7)$$

$$k=60⟹59,61\text{(Both are prime!)}$$

$$k=70⟹69,71(69=3\times 23)$$

$$k=80⟹79,81(81=9\times 9)$$

$$k=90⟹89,91(91=7\times 13)$$

$$k=100⟹99,101(99=9\times 11)$$

Step 6: Conclude the final count

Only $k=60$ produces a valid pair of primes. This corresponds to exactly one four-digit number.

$$k=60⟹N=60^2-1=3599$$

Correct answer: B. $1$
There is exactly 1 such number (3599). This matches answer choice B. Final Check: 3599 ends in 99, 3599 + 1 = 3600 = 60^2, and 3599 = 59 * 61, where both 59 and 61 are prime. All conditions are perfectly satisfied.

Solution 2

Twin Prime Divisibility Shortcut

We can bypass testing most of the cases by observing that $k-1$, $k$, and $k+1$ are three consecutive integers. Because $k-1$ and $k+1$ must be twin primes greater than 3, $k$ must be a multiple of 3.

Divisibility of Consecutive Integers: Among any three consecutive integers, exactly one must be a multiple of 3. We can use this to quickly eliminate candidates.

Step 1: Set up the required conditions for $k$

From our standard method, we know $k$ must be a multiple of 10 between 40 and 100. We also established that $k-1$ and $k+1$ must both be prime numbers.

$$k\in \{40,50,60,70,80,90,100\}$$

Step 2: Analyze the divisibility of three consecutive integers

Consider the three consecutive integers centered at $k$. According to fundamental divisibility rules, exactly one of these three consecutive integers must be a multiple of 3.

$$(k-1),k,(k+1)$$

Step 3: Deduce that $k$ is a multiple of 3

We need $k-1$ and $k+1$ to be prime numbers. Since the smallest candidate for $k-1$ is 39, both prime numbers would be much larger than 3. A prime number strictly greater than 3 cannot be a multiple of 3. Therefore, neither $k-1$ nor $k+1$ is the multiple of 3. The multiple of 3 must be $k$ itself.

$$k\equiv 0(\mathrm{mod}3)$$

Step 4: Combine divisibility rules to limit candidates

We already know that $k$ is a multiple of 10. Since 3 and 10 share no common factors, $k$ must be a multiple of their product, 30. Looking at our list of candidates, only two numbers fit this strict requirement.

$$k\equiv 0(\mathrm{mod}30)$$

$$k\in \{60,90\}$$

Step 5: Test the remaining candidates

By applying this logic, we bypassed five unnecessary tests. We only need to check $k=60$ and $k=90$. We find that $k=60$ yields two primes, while $k=90$ yields 91, which is composite. This leaves exactly one valid number.

$$k=60⟹59,61\text{(Both are prime)}$$

$$k=90⟹89,91(91=7\times 13)$$

Correct answer: B. $1$
We confidently conclude there is 1 valid number (choice B). The divisibility shortcut ensures no other cases could possibly yield two prime factors.

Similar Problems in Real Exams

The following five AMC 8 problems are closely related to AMC 8 2025 Question 23. These real AMC 8 problems test the same key concept—Primes & Prime Factorization. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2023 AMC 8 Problem 4

2016 AMC 8 Problem 9

2014 AMC 8 Problem 4

2014 AMC 8 Problem 23

2012 AMC 8 Problem 18

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Difference of Squares Factorization:

$a$ and $b$ are any numbers or algebraic terms. It is used to factor $N=k^2-1$ into $(k-1)(k+1)$, directly exposing the potential prime factors of $N$.

• Unique Divisors of a Product of Two Primes:

$p$ and $q$ are prime numbers. It proves that the integer factors $(k-1)$ and $(k+1)$ must be exactly the primes $p$ and $q$, since neither algebraic factor can be 1 when $k\ge 40$.

• Divisibility of Consecutive Integers:

$n$ is a positive integer (we used $n=3$). It establishes that one of the numbers $(k-1),k,(k+1)$ must be divisible by 3, which is the core logical engine of the shortcut method.

Key Takeaway: Remember This Strategy for Next Time

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