Primes

What you need to know

• A prime is an integer $p > 1$ whose only positive divisors are $1$ and $p$. The integer $1$ is not prime.
• $2$ is the only even prime; $3$ is the only prime divisible by $3$; $5$ is the only prime ending in $5$. Every prime $p > 3$ satisfies $p \equiv \pm 1 \pmod 6$.
• To test whether $n$ is prime, trial-divide by primes up to $\sqrt{n}$. For $n < 289$ it suffices to check $2, 3, 5, 7, 11, 13$.
• Primes below $100$: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97$ (25 in all).
• If a prime $p$ divides a product $ab$, then $p \mid a$ or $p \mid b$.

How AMC 10 tests it

• A quadratic whose roots are both prime, with the sum or product given: an odd sum forces one root to be $2$.
• "For how many primes $p$ is $p^2 + c$ (or $p + a$ and $p + b$) also prime?" The answer is almost always $0$ or $1$, decided by a small modulus.
• Arrange digits into two-digit primes, or count primes with a digit condition; this rewards knowing the list below $100$ cold.
• Problems on the smallest prime factor of a large expression such as $2^{20} - 1$ or $n! + 1$, using algebraic factorization.
• These appear at problems 6–18 and test speed and parity awareness rather than deep theory.

Standard approaches

1. Handle $2$ (and often $3$) separately; then every remaining prime is odd and not a multiple of $3$.
2. Reduce modulo $3$, $4$, or $6$: show every case except one makes the expression divisible by a small prime and larger than it.
3. Factor algebraically ($a^2 - b^2$, sum of cubes, $x^4 + 4y^4$) before hunting for prime divisors.
4. Use Vieta's formulas to turn "prime roots" into a sum-and-product condition on two primes.
5. When the count is small, just list; know the primes below $100$ cold.

Worked example

How many primes $p$ have the property that both $p + 10$ and $p + 14$ are also prime?

(A) $0$ (B) $1$ (C) $2$ (D) $3$ (E) infinitely many

Consider residues modulo $3$. Since $10 \equiv 1$ and $14 \equiv 2 \pmod 3$, the three numbers $p$, $p+10$, $p+14$ are congruent to $p$, $p+1$, $p+2$, which cover all three residues. So one of them is divisible by $3$. A prime divisible by $3$ must equal $3$; since $p + 10 > 3$ and $p + 14 > 3$, the only possibility is $p = 3$. Check: $13$ and $17$ are prime. Exactly one prime works, so the answer is $\boxed{\textbf{(B)}\ 1}$.

Pitfalls

• Treating $1$ as prime, or forgetting that $2$ is prime when counting.
• Stopping trial division too early, or too late: check primes up to $\sqrt{n}$ only.
• Concluding "divisible by $3$" means composite without confirming the number exceeds $3$.
• Misremembering the list: $91 = 7 \cdot 13$ and $57 = 3 \cdot 19$ are not prime, while $97$ and $89$ are.

Traps that recur

• Taking b_1 = 60 (so 61!/60! = 61) without checking whether a smaller b_1 still lets the leftover primes cancel. (2013 AMC 10B #20)
• Missing that the two factors must have the same parity, and then wasting time on pairs like 1 x 48 or 3 x 16, or accepting q = 8 or p = 1 as prime. (2005 AMC 10A #24)
• Stopping at the primes 71 and 73 and reporting a digit sum of one of them (8 or 10), instead of the next prime 79. (2022 AMC 10B #13)
• Testing only p = 2 and p = 3 and picking the first expression that looks composite, without checking the other choices produce a prime for some p. (2018 AMC 10B #11)