Primes

What you need to know

• A prime is a whole number greater than $1$ whose only positive divisors are $1$ and itself. The number $1$ is not prime. Example: $13$ is prime; $15 = 3\cdot5$ is not.
• The small primes are special. $2$ is the only even prime, $3$ is the only prime divisible by $3$, and $5$ is the only prime ending in $5$.
• Primes above $3$ sit next to multiples of $6$. Every prime $p > 3$ leaves remainder $1$ or $5$ when divided by $6$, written $p \equiv \pm1 \pmod 6$. Any other remainder ($0$, $2$, $3$, $4$) would make $p$ divisible by $2$ or $3$. Example: $29 = 30-1$ and $31 = 30+1$. It only works one way: $25 = 24+1$ is not prime.
• Testing whether $n$ is prime. Divide by the primes up to $\sqrt n$; if none divides $n$, it is prime. (If $n = ab$ with both $a$ and $b$ bigger than $\sqrt n$, then $ab$ would be bigger than $n$.) So for $n < 289 = 17^2$, checking $2, 3, 5, 7, 11, 13$ is enough. Example: testing $221$, you reach $13$ and find $221 = 13\cdot17$.
• Primes below $100$ (there are $25$): $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97$.
• If a prime $p$ divides a product $ab$, then $p$ divides $a$ or $p$ divides $b$. This fails for numbers that aren't prime. Example: $6$ divides $4\cdot9 = 36$ but divides neither $4$ nor $9$.

How AMC 10 tests it

• A quadratic whose roots are both prime, with their sum or product given. An odd sum forces one root to be $2$, since two odd primes add to an even number.
• "For how many primes $p$ is $p^2+c$ (or both $p+a$ and $p+b$) also prime?" The answer is almost always $0$ or $1$, decided by a small modulus.
• Arranging digits into two-digit primes, or counting primes with a digit condition. This rewards knowing the primes below $100$ by heart.
• The smallest prime factor of a large expression such as $2^{20}-1$ or $n!+1$, using algebraic factoring.
• These show up as problems 6–18 and test speed and attention to odd and even rather than deep theory.

Standard approaches

1. Handle $2$ (and often $3$) separately; every prime left is odd and not a multiple of $3$.
2. Work mod $3$, $4$, or $6$. Show that in every case but one, the expression is divisible by a small prime and bigger than it, so it is not prime.
3. Factor with algebra before hunting for prime divisors: $a^2-b^2 = (a-b)(a+b)$, the sum of cubes $a^3+b^3 = (a+b)(a^2-ab+b^2)$, and $x^4+4y^4 = (x^2+2xy+2y^2)(x^2-2xy+2y^2)$. Example: $2^{20}-1 = (2^{10}-1)(2^{10}+1) = 1023\cdot1025$, and $1023 = 3\cdot341$, so the smallest prime factor is $3$.
4. Use Vieta's Formulas to turn "prime roots" into a sum-and-product condition on two primes.
5. When the count is small, list the cases. Know the primes below $100$ by heart.

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

Step 1: remainders mod $3$. $10$ leaves remainder $1$ and $14$ leaves remainder $2$ when divided by $3$. So $p$, $p+10$, $p+14$ leave the same remainders as $p$, $p+1$, $p+2$.

Step 2: one is a multiple of $3$. One of any three integers in a row is a multiple of $3$, so one of $p$, $p+10$, $p+14$ is divisible by $3$. For example, $p = 7$ gives $7$, $17$, $21$, and $21 = 3\cdot7$.

Step 3: it must be $3$ itself. The only prime divisible by $3$ is $3$. Since $p+10$ and $p+14$ are bigger than $3$, the multiple of $3$ must be $p$, so $p = 3$.

Step 4: check. $13$ and $17$ are both prime, so exactly one prime works.

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 going past $\sqrt n$. Example: stopping at $7$ would wrongly call $187$ prime, but $187 = 11\cdot17$. To test $97$, you can stop at $7$, since $11^2 = 121 > 97$.
• Deciding a number divisible by $3$ is not prime without checking that it is bigger than $3$. That is exactly why $p = 3$ works above.
• Misremembering the list: $91 = 7\cdot13$ and $57 = 3\cdot19$ 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)