Primes · Lecture 15 · nt-primes
📺 Course materials — Lecture 15:
lecture video · handout PDF (Lecture 15) · board notes (15.jpg) · homework answer key (Chapter 15)
— all in the
amc8/video folder.
·
🖨 Printable PDF (answers & solutions at the back)
🔑 Key facts
A prime has exactly two factors: $1$ and itself. A composite number has three or more.
$0$ and $1$ are neither prime nor composite — the instructor circles this every year.
Every integer $\ge 2$ breaks into primes in exactly one way (e.g. $12 = 2^2\times3$) — that factorization is
the number's "ID card", and almost every problem in this lecture starts by reading it.
There are $25$ primes below $100$; the first few are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, \ldots$
🧰 Instructor's toolbox (from the lecture video & board notes)
1 · Classify by counting factors.
$13 = 1\times13$ and nothing else — two factors, prime. $12 = 1\times12 = 2\times6 = 3\times4$ — six factors, composite.
A composite number always has at least three factors. And $0, 1$ are neither prime nor composite —
answer-choice writers love to bait you with "$1$ is prime".
2 · Prime fingerprints. $2$ is the only even prime; $5$ is the only prime ending in $5$;
every other prime ends in $1, 3, 7$ or $9$. So when a problem says two primes add up to an odd number,
one of them must be $2$ (odd $=$ odd $+$ even, and the even prime is unique) — that single observation solves Example 1.
3 · Test a prime by trial division — but only up to $\sqrt{N}$.
Try the primes $2, 3, 5, 7,\ldots$ in order; you may stop as soon as the square of the next prime exceeds $N$.
Is $101$ prime? Test $2,3,5,7$ and stop ($11^2 = 121 > 101$) — prime.
Is $503$ prime? Test primes up to $19$ ($23^2 = 529 > 503$) — prime.
Is $2023$ prime? $45^2 = 2025$, so test primes up to $43$… and $2023 = 7\times17^2$ — composite. The board's warning:
skipping the square-root cutoff wastes minutes you don't have.
5 · Squares and cubes wear their exponents.
A number is a perfect square exactly when every exponent in its prime factorization is even
($2^4\!\times\!5^2\!\times\!7^6$ ✓, $\;2^2\!\times\!5\!\times\!7^3$ ✗), and a perfect cube exactly when every exponent is a
multiple of 3. So $4^{2019} = 2^{4038}$ is a square while $2^{2017}$ is not (Practice 6), and "make $360x$ a square"
just means "top up each exponent to an even number" (Practice 7).
Keep $2^{10}=1024$ and the cubes $1,8,27,64,125,216,343$ in your pocket.
6 · The divisor-counting theorem. If $N = p^a q^b r^c\cdots$, the number of positive divisors is
\[d(N) = (a+1)(b+1)(c+1)\cdots\]
— multiply the (exponent$+1$)'s, never add. And memorize the tiny table:
exactly $1$ factor $\to$ the number $1$; exactly $2$ $\to$ a prime; exactly $3$ $\to$ a prime squared ($4,9,25,49,\ldots$).
This runs Example 5 and Practice 11, 13 — and the whole Counting Divisors handout.
8 · Parity is the fastest filter.
odd $\pm$ odd $=$ even; odd $\pm$ even $=$ odd; odd $\times$ odd $=$ odd; anything $\times$ even $=$ even.
Also $m^2$ has the same parity as $m$. Run the options through the parity filter before doing any real work —
Practice 5, 9, 10 fall to it alone.
9 · Consecutive integers: use the middle.
An odd count of consecutive integers sums to (count) $\times$ (middle term): the board's "$x-2d,\,x-d,\,x,\,x+d,\,x+2d$
adds to $5x$". Five consecutive integers summing to $35$? The middle is $7$ — no equations needed (Practice 3).
📖 Worked examples
2014 AMC 8 #4 · only even prime
Example 1. The sum of two prime numbers is $85$. What is the product of these two prime numbers?
(A) 85(B) 91(C) 115(D) 133(E) 166
Key idea: $85$ is odd, and odd $=$ odd $+$ even — so one of the two primes is even.
The only even prime is $2$, forcing the other to be $85 - 2 = 83$ (prime ✓).
Product: $2 \times 83 = 166$.
Answer: E
2012 AMC 8 #13 · prime factorization
Example 2. Jamar bought some pencils costing more than a penny each at the school bookstore and paid $\$1.43$.
Sharona bought some of the same pencils and paid $\$1.87$. How many more pencils did Sharona buy than Jamar?
(A) 2(B) 3(C) 4(D) 5(E) 6
Key idea: the unit price (in cents, an integer $>1$) divides both totals, so factor them:
$143 = 11\times13$ and $187 = 11\times17$. The only shared factor bigger than $1$ is $11$ — the price per pencil.
So Jamar bought $13$, Sharona bought $17$, and $17-13 = 4$.
Answer: C
2018 AMC 8 #14 · digit product
Example 3. Let $N$ be the greatest five-digit number whose digits have a product of $120$. What is the sum of the digits of $N$?
(A) 15(B) 16(C) 17(D) 18(E) 20
Key idea: $120 = 2^3\times3\times5$, and each digit must be $1$–$9$.
To make $N$ as large as possible, make its
leading digit as large as possible: the biggest single digit you can
assemble from $2,2,2,3,5$ is $8 = 2^3$. That leaves $5$ and $3$, and pad with $1$s: $120 = 8\times5\times3\times1\times1$,
so $N = 85311$. Digit sum: $8+5+3+1+1 = 18$.
Answer: D
2016 AMC 8 #15 · difference of squares
Example 4. What is the largest power of $2$ that is a divisor of $13^4 - 11^4$?
(A) 8(B) 16(C) 32(D) 64(E) 128
Key idea: never compute $13^4$. Factor with the difference of squares, twice:
\[13^4-11^4 = (13^2+11^2)(13^2-11^2) = (169+121)(13+11)(13-11) = 290\times24\times2.\]
Now read off the $2$s: $290 = 2\cdot5\cdot29$, $\;24 = 2^3\cdot3$, and the last factor is $2$ —
five twos in all, so the largest power of $2$ dividing the number is $2^5 = 32$.
Answer: C
2020 AMC 8 #17 · divisor counting
Example 5. How many factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1, 2, 3, 4, 6,$ and $12$.)
(A) 6(B) 7(C) 8(D) 9(E) 10
Key idea: $2020 = 2^2\times5\times101$, so it has $(2+1)(1+1)(1+1) = 12$ divisors.
Counting the ones with
more than $3$ factors directly is risky — count the opposite instead.
"$3$ or fewer factors" means: the number $1$, a prime, or a prime squared. In the divisor list that's
$1$; the primes $2, 5, 101$; and $4 = 2^2$ — five of them. Everything else qualifies: $12 - 5 = 7$.
Answer: B
✏️ Practice
Try each one before opening the solution.
2010 AMC 8 #14
1. What is the sum of the prime factors of $2010$?
(A) 67(B) 75(C) 77(D) 201(E) 210
Solution
Peel primes: $2010 = 2\times3\times5\times67$. Sum: $2+3+5+67 = 77$.
Answer: C
2014 AMC 8 #23
2. Three members of the Euclid Middle School girls' softball team had the following conversation.
Ashley: I just realized that our uniform numbers are all $2$-digit primes.
Bethany: And the sum of your two uniform numbers is the date of my birthday earlier this month.
Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month.
Ashley: And the sum of your two uniform numbers is today's date.
What number does Caitlin wear?
(A) 11(B) 13(C) 17(D) 19(E) 23
Solution
Every pairwise sum is a date, so each is at most $31$ — the three primes must come from the smallest
two-digit primes $11, 13, 17, 19$, and $19$ fails ($19+13=32>31$). So the numbers are $11, 13, 17$ with pairwise sums
$24, 28, 30$. Bethany's birthday (Ashley $+$ Caitlin) is the
earliest date and today (Bethany $+$ Caitlin) is
in between, so Caitlin $<$ Ashley $<$ Bethany — Caitlin wears the smallest number, $11$.
Answer: A
2015 AMC 8 #23
3. Tom has twelve slips of paper which he wants to put into five cups labeled $A$, $B$, $C$, $D$, $E$. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from $A$ to $E$. The numbers on the papers are $2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,$ and $4.5$. If a slip with $2$ goes into cup $E$ and a slip with $3$ goes into cup $B$, then the slip with $3.5$ must go into what cup?
(A) $A$(B) $B$(C) $C$(D) $D$(E) $E$
Solution
All the slips together add to $35$, and five consecutive integers summing to $35$ have middle term
$35/5 = 7$: the cup sums are $5, 6, 7, 8, 9$. Now place the $3.5$: cup $A$ (sum $5$) would need a $1.5$ — no such slip;
cup $B$ already has $3$, and $3+3.5 = 6.5 > 6$; cup $C$ would need the rest to add to $3.5$ — impossible with the slips left;
cup $E$ has $2$, needing another $3.5$ beyond the one we placed — impossible. Cup $D$ works: $3.5 + 2 + 2.5 = 8$ ✓.
Answer: D
2019 AMC 8 #19
4. In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?
(A) 22(B) 23(C) 24(D) 26(E) 30
Solution
Let each top team beat all three bottom teams twice: $6$ wins $= 18$ points each.
Among themselves the three top teams play $4$ games each and must stay tied — the best balanced split is
win one, lose one against each rival: $2$ more wins $= 6$ points. Total $18 + 6 = 24$.
(All $30$ points would require beating the other top teams too — then they're not tied.)
Answer: C
2010 AMC 8 #22
5. The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
(A) 0(B) 2(C) 4(D) 6(E) 8
Solution
Write the number as $100a + 10b + c$ with $a = c + 2$. Reversing gives $100c + 10b + a$, and
\[(100a+10b+c)-(100c+10b+a) = 99(a-c) = 99\times2 = 198.\]
The difference is
always $198$ — units digit $8$.
Answer: E
2016 AMC 8 #7
6. Which of the following numbers is not a perfect square?
(A) $1^{2016}$(B) $2^{2017}$(C) $3^{2018}$(D) $4^{2019}$(E) $5^{2020}$
Solution
A power is a square when it can be written with an even exponent:
$1^{2016}=1$, $3^{2018}=(3^{1009})^2$, $4^{2019}=2^{4038}=(2^{2019})^2$, $5^{2020}=(5^{1010})^2$.
But $2^{2017}$ has an odd exponent on the prime $2$ — not a square.
Answer: B
2009 AMC 8 #17
7. The positive integers $x$ and $y$ are the two smallest positive integers for which the product of $360$ and $x$ is a square and the product of $360$ and $y$ is a cube. What is the sum of $x$ and $y$?
(A) 80(B) 85(C) 115(D) 165(E) 610
Solution
$360 = 2^3\times3^2\times5$. For a
square, every exponent must become even:
top up $2^3 \to 2^4$ and $5 \to 5^2$, so $x = 2\times5 = 10$. For a
cube, every exponent must reach a multiple of $3$:
$3^2 \to 3^3$ and $5 \to 5^3$, so $y = 3\times5^2 = 75$. Then $x + y = 85$.
Answer: B
2018 AMC 8 #25
8. How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive?
(A) 4(B) 9(C) 10(D) 57(E) 58
Solution
$2^8+1 = 257$, and $6^3 = 216 < 257 \le 343 = 7^3$, so the first cube in range is $7^3$.
At the top, $2^{18} = (2^6)^3 = 64^3$, so $64^3 < 2^{18}+1$ is in range — the last cube is $64^3$.
Count: $64 - 7 + 1 = 58$.
Answer: E
2005 AMC 8 #8
9. Suppose $m$ and $n$ are positive odd integers. Which of the following must also be an odd integer?
(A) $m+3n$(B) $3m-n$(C) $3m^2+3n^2$(D) $(nm+3)^2$(E) $3mn$
Solution
Run the parity filter: $m + 3n$ is odd$+$odd $=$ even; $3m - n$ likewise even;
$3m^2 + 3n^2$ is odd$+$odd $=$ even; $nm + 3$ is odd$+$odd $=$ even, so its square is even.
Only $3mn$ — a product of three odds — stays odd.
Answer: E
2014 AMC 8 #13
10. If $n$ and $m$ are integers and $n^2 + m^2$ is even, which of the following is impossible?
(A) $n$ and $m$ are even(B) $n$ and $m$ are odd(C) $n+m$ is even(D) $n+m$ is odd(E) none of these are impossible
Solution
$m^2$ and $n^2$ have the same parity as $m$ and $n$. For the sum of the squares to be even,
$m$ and $n$ are
both even or
both odd — and in either case $n+m$ is even. So $n+m$ odd is impossible.
Answer: D
2018 AMC 8 #18
11. How many positive factors does $23{,}232$ have?
(A) 9(B) 12(C) 28(D) 36(E) 42
Solution
Peel primes: $23232 = 2^6\times3\times11^2$.
Divisor count: $(6+1)(1+1)(2+1) = 7\times2\times3 = 42$.
Answer: E
2017 AMC 8 #23
12. Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by $5$ minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?
(A) 10(B) 15(C) 25(D) 50(E) 82
Solution
If one mile takes $x$ minutes, the day's distance is $60/x$ miles — so $x, x+5, x+10, x+15$ must
all divide $60$. Scanning the divisors of $60$: only $5, 10, 15, 20$ works ($x=5$).
Distances: $12 + 6 + 4 + 3 = 25$ miles.
Answer: C
2015 AMC 8 #22
13. On June $1$, a group of students are standing in rows, with $15$ students in each row. On June $2$, the same group is standing with all of the students in one long row. On June $3$, the same group is standing with just one student in each row. On June $4$, the same group is standing with $6$ students in each row. This process continues through June $12$ with a different number of students per row each day. However, on June $13$, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?
(A) 21(B) 30(C) 60(D) 90(E) 1080
Solution
Twelve days of different row sizes means the total has exactly $12$ divisors.
Check the choices with the theorem: $21 = 3\times7$ gives $4$; $30 = 2\times3\times5$ gives $8$;
$60 = 2^2\times3\times5$ gives $(2+1)(1+1)(1+1) = 12$ ✓ — and it's the smallest choice that works
(both $15$ and $6$ do divide $60$ ✓).
Answer: C
mock · course problem
14. How many $3$-digit palindromes are there? (A palindrome reads the same left-to-right and right-to-left, like $121$ or $737$.)
(A) 30(B) 60(C) 90(D) 120(E) 180
Solution
A $3$-digit palindrome is $\overline{aba}$: the first and last digits match and can't be $0$ —
$9$ choices — and the middle digit is free — $10$ choices. $9\times10 = 90$.
Answer: C
2019 AMC 8 #13
15. A palindrome is a number that has the same value when read from left to right or from right to left. (For example, $12321$ is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$?
(A) 2(B) 3(C) 4(D) 5(E) 6
Solution
Two-digit palindromes are exactly $11, 22, \ldots, 99$ — all multiples of $11$, so the sum of three of them
is a multiple of $11$ too. Hunting upward from $100$: the multiples $110 = 11+22+77$ works and isn't a palindrome ✓
(the smaller sums like $66, 77,\ldots, 99$ are palindromes or too small). So $N = 110$ and its digit sum is $1+1+0 = 2$.
Answer: A
🆕 Fresh from the latest exams
The same ideas, exactly as they appeared on the most recent tests.
2024 AMC 8 #4
16. When Yunji added all the integers from $1$ to $9$, she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?
(A) 5(B) 6(C) 7(D) 8(E) 9
Solution
$1+2+\cdots+9 = 45$. Leaving out $x$ gives $45 - x$, which must be a perfect square with
$36 \le 45-x \le 44$ — only $36$ fits, so $x = 9$.
Answer: E
2025 AMC 8 #23
17. How many four-digit numbers have all three of the following properties? (I) The tens and ones digit are both $9$. (II) The number is $1$ less than a perfect square. (III) The number is the product of exactly two prime numbers.
(A) 0(B) 1(C) 2(D) 3(E) 4
Solution
If $N+1 = k^2$ and $N$ ends in $99$, then $k^2$ ends in $00$, so $k$ is a multiple of $10$.
Four-digit range: $k = 40, 50, 60, 70, 80, 90$. Factor $N = k^2 - 1 = (k-1)(k+1)$ — it's a product of exactly two primes
only when
both neighbors are prime: $39\times41$ ✗, $49\times51$ ✗, $59\times61$ ✓, $69\times71$ ✗, $79\times81$ ✗,
$89\times91$ ✗ ($91 = 7\times13$!). Just one: $3599 = 59\times61$.
Answer: B
2026 AMC 8 #18
18. In how many ways can $60$ be written as the sum of two or more consecutive odd positive integers that are arranged in increasing order?
(A) 1(B) 2(C) 3(D) 4(E) 5
Solution
$j$ consecutive odd numbers starting at $a$ add to $j(a + j - 1)$, so we need $j(a+j-1) = 60$
with $j \ge 2$, $a$ odd and positive. For each divisor $j$ of $60$: $a = \tfrac{60}{j} - j + 1$ must be odd and $\ge 1$.
$j=2: a=29$ ✓ ($29+31$); $\;j=3: a=18$ ✗; $\;j=4: a=12$ ✗; $\;j=5: a=8$ ✗; $\;j=6: a=5$ ✓ ($5+7+9+11+13+15$);
$\;j=10: a<0$ ✗. Two ways.
Answer: B
🗝 Answer key
| Example | 1 | 2 | 3 | 4 | 5 |
|---|
| E | C | D | C | B |
|---|
| Practice | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
|---|
| C | A | D | C | E | B | B | E | E | D | E | C | C | C | A | E | B | B |
|---|
Sources: prep-course Lecture 15 (video + board notes + 20 worked examples, translated to English) and official AMC 8 problems from the local bank
(kb/content/banks/). Answers verified against the bank's official answer keys.
Companion handout: Counting Divisors.