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Primes

Prime properties, prime factorization, counting with primes

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PrimesAMC 10 Step by Step

Name: Date: October 28, 2026

17 problems, easiest first

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

Handle $2$ (and often $3$ ) separately; every prime left is odd and not a multiple of $3$ .
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.
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 $3$ .
Use Vieta's Formulas to turn "prime roots" into a sum-and-product condition on two primes.
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)

1.2024 AMC 10A #5
What is the least value of $n$ such that $n!$ is a multiple of $2024$ ?
1. A) $11$
2. B) $21$
3. C) $22$
4. D) $23$
5. E) $253$
2.2020 AMC 10B #4
The acute angles of a right triangle are $a^{\circ}$ and $b^{\circ}$ , where $a>b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$ ?
1. A) $2$
2. B) $3$
3. C) $5$
4. D) $7$
5. E) $11$
3.2024 AMC 10A #3
What is the sum of the digits of the smallest prime that can be written as a sum of $5$ distinct primes?
1. A) $5$
2. B) $7$
3. C) $8$
4. D) $10$
5. E) $13$
4.2022 AMC 10B #6
How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
1. A) $0$
2. B) $1$
3. C) $2$
4. D) $3$
5. E) $4$
5.2021 AMC Fall 10B #8
The greatest prime number that is a divisor of $16{,}384$ is $2$ because $16{,}384 = 2^{14}$ . What is the sum of the digits of the greatest prime number that is a divisor of $16{,}383$ ?
1. A) $3$
2. B) $7$
3. C) $10$
4. D) $16$
5. E) $22$
6.2021 AMC Fall 10A #5
The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
1. A) $1$
2. B) $3$
3. C) $5$
4. D) $7$
5. E) $9$
7.2015 AMC 10B #11
Among the positive integers less than $100$ , each of whose digits is a prime number, one is selected at random. What is the probability that the selected number is prime?
1. A) $\dfrac{8}{99}$
2. B) $\dfrac{2}{5}$
3. C) $\dfrac{9}{20}$
4. D) $\dfrac{1}{2}$
5. E) $\dfrac{9}{16}$
8.2002 AMC 10B #6
For how many positive integers $n$ is $n^2 - 3n + 2$ a prime number?
1. A) $\text{none}$
2. B) $\text{one}$
3. C) $\text{two}$
4. D) $\text{more than two, but finitely many}$
5. E) $\text{infinitely many}$
9.2002 AMC 10A #15
Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?
1. A) $150$
2. B) $160$
3. C) $170$
4. D) $180$
5. E) $190$
10.2000 AMC 10 #11
Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
1. A) $22$
2. B) $60$
3. C) $119$
4. D) $180$
5. E) $231$
11.2022 AMC 10B #13
The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
1. A) $8$
2. B) $10$
3. C) $11$
4. D) $13$
5. E) $16$
12.2018 AMC 10B #11
Which of the following expressions is never a prime number when $p$ is a prime number?
1. A) $p^2+16$
2. B) $p^2+24$
3. C) $p^2+26$
4. D) $p^2+46$
5. E) $p^2+96$
13.2003 AMC 10A #14
Let $n$ be the largest integer that is the product of exactly 3 distinct prime numbers $d$ , $e$ , and $10d+e$ , where $d$ and $e$ are single digits. What is the sum of the digits of $n$ ?
1. A) $12$
2. B) $15$
3. C) $18$
4. D) $21$
5. E) $24$
14.2002 AMC 10B #15
The positive integers $A, B, A-B,$ and $A+B$ are all prime numbers. The sum of these four primes is
1. A) $\mathrm{even}$
2. B) $\mathrm{divisible by }3$
3. C) $\mathrm{divisible by }5$
4. D) $\mathrm{divisible by }7$
5. E) $\mathrm{prime}$
15.2002 AMC 10A #14
Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is
1. A) $0$
2. B) $1$
3. C) $2$
4. D) $4$
5. E)more than 4
16.2013 AMC 10B #20
The number $2013$ is expressed in the form $2013=\frac{a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!},$ where $a_1\ge a_2\ge\cdots\ge a_m$ and $b_1\ge b_2\ge\cdots\ge b_n$ are positive integers and $a_1+b_1$ is as small as possible. What is $|a_1-b_1|$ ?
1. A) $1$
2. B) $2$
3. C) $3$
4. D) $4$
5. E) $5$
17.2005 AMC 10A #24
For each positive integer $m > 1$ , let $P(m)$ denote the greatest prime factor of $m$ . For how many positive integers $n$ is it true that both $P(n) = \sqrt{n}$ and $P(n+48) = \sqrt{n+48}$ ?
1. A) $0$
2. B) $1$
3. C) $3$
4. D) $4$
5. E) $5$

Primes — answersAMC 10 Step by Step

Name: Date: October 28, 2026

Keep this page separate from the problems.

Answers

#	Answer	The idea it turns on
1	D	Break 2024 into primes: 2 times 2 times 2 times 11 times 23. The factorial must include the prime 23, and 23! already includes everything else.
2	D	The two acute angles add to 90. Try b = 2, 3, 5, 7 in order and stop at the first one where 90 minus b is also prime.
3	B	If 2 is one of the five primes, the total is even and cannot be prime. So use five odd primes, and try the smallest totals first.
4	A	Every term is a product: 121 = 11 × 11, 11211 = 111 × 101, 1112111 = 1111 × 1001. The pattern never stops, so no term is prime.
5	C	16383 is 128 squared minus 1, so it factors as (128 - 1)(128 + 1) = 127 times 129 = 3 times 43 times 127, and 127 is prime.
6	E	Cross out answers with quick tests: 5 ends in 5, digits 1 and 7 make the digit sum a multiple of 3, and 202103 is a multiple of 11.
7	B	Only the digits 2, 3, 5, 7 are allowed, so there are just 4 + 16 = 20 numbers to check. Eight of them are prime.
8	B	The expression factors as (n - 1)(n - 2), two whole numbers next to each other, so it is always even. The only even prime is 2.
9	E	A two-digit prime cannot end in an even digit or in 5, so 1, 3, 7, 9 must be the last digits and 2, 4, 5, 6 the first digits.
10	C	Both primes are odd, so the product is odd and the sum is even, and the result is odd. The largest result is 191, which leaves only 119.
11	E	Call the primes p and p + 2. Expanding the cubes leads to (p + 1)² = 5184 = 72², so the primes are 71 and 73.
12	C	The square of any prime other than 3 is one more than a multiple of 3, so adding 26 always gives a multiple of 3. The prime 3 gives 35.
13	A	d and e must be among 2, 3, 5, 7, and a two-digit prime cannot end in 2 or 5. That leaves only a few cases to list and compare.
14	E	A-B and A+B are both odd primes, so B must be the even prime 2. Then A-2, A, A+2 are all prime, which only 3, 5, 7 manage.
15	B	The two roots add to 63, an odd number. Two odd primes would add to an even number, so one root must be 2 and the other 61.
16	B	2013 has the prime factor 61, so the top needs 61!. That brings in the prime 59, so the bottom needs 59! to cancel it.
17	B	The largest prime factor equals the square root only for a prime squared. So look for two prime squares 48 apart: 121 and 169.

Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.