AMC 8 Wiki › Algebra & Number Theory › Diophantine Equations

Diophantine Equations · Lecture 16 · alg-diophantine

📺 Course materials — Lecture 16: lecture video · handout PDF (Lecture 16) · board notes (16.jpg) · homework answer key (Chapter 16) — all in the amc8/video folder.  ·  🖨 Printable PDF (answers & solutions at the back)

🔑 Key facts

A Diophantine equation has more unknowns than equations — like $3x+y=28$ — but demands integer (often positive-integer) solutions. Integrality is the missing equation: it shrinks "infinitely many solutions" down to a handful you can list.

For $ax+by=c$, once you have one solution, all the others step by the opposite coefficients: $x$ changes by $b$ while $y$ changes by $-a$. Counting solutions = counting the steps that stay inside the allowed range.

Partner facts for the LCM/GCF half of the lecture: $\;\gcd(A,B)\cdot\mathrm{lcm}(A,B)=A\cdot B$, and in prime factorizations GCF takes the smaller exponent of each prime, LCM the larger.

🧰 Instructor's toolbox (from the lecture video & board notes)

1 · Name the unknowns and write the equation — even if there's only one. "One positive integer is ___ more than twice another, and they sum to $28$" becomes $2x+y+x=28$, i.e. $3x+y=28$. Don't panic that one equation holds two unknowns: with integer constraints, that's a solvable problem, not a broken one. (Example 2 lives entirely on this line.)
2 · Parity check before any work. $2x+4y=41$: the left side is even + even, the right side is odd — no solutions, done in five seconds. Always glance at even/odd (or a quick mod) before you start enumerating; the instructor calls this the free win.
3 · Push divisibility through the equation. In $2x+3y=45$: since $3\mid 45$ and $3\mid 3y$, subtracting gives $3\mid 2x$, so $3\mid x$. Now enumerate only $x=0,3,6,\dots$ with $y=15,13,11,\dots$ Stronger version: take the whole equation mod a coefficient. $5x+6y=49 \pmod 5$: $6y\equiv y$ and $49\equiv 4$, so $y\equiv 4\pmod 5$ — try $y=4$ and $x=5$ pops out.
4 · One solution, then step by coefficients. $5x+y=50$: spot $(x,y)=(0,50)$; each $+1$ in $x$ costs $-5$ in $y$: $(0,50)\to(1,45)\to(2,40)\to\cdots$ In general $x=k,\;y=50-5k$. Written from the other end: $x=10-k,\;y=5k$. Same ladder, two starting rungs.
(0, 50) (1, 45) (2, 40) x+1, y−5 x+1, y−5 in ax+by=c the steps are x→x+b, y→y−a
5 · Count solutions with the bounds. From $y=50-5k\ge 0$: $k\le 10$, so $k=0,1,\dots,10$ — $11$ solutions. Want only $30\le y\le 40$? Then $30\le 50-5k\le 40 \Rightarrow -20\le -5k\le -10 \Rightarrow 2\le k\le 4$ — $3$ solutions. Careful: dividing by a negative flips the inequality — the board circles that flip in red.
6 · Two equations, three unknowns → eliminate down to one Diophantine line. The socks system $x+y+z=12$, $x+3y+4z=24$: subtract to kill $x$, leaving $2y+3z=12$; then bounds + parity finish it (Example 3). Getting rid of one variable turns a scary system into tip #4.
7 · Fractions of an unknown total → the total is a multiple of the LCM of the denominators. If $\tfrac13$ of the marbles are blue and $\tfrac14$ are red, the total is a multiple of $12$ (Example 1). If $\tfrac14 x + \tfrac27 x$ must be whole and bounded, $x$ moves in steps of $28$ — that plus a bound pins $x=56$ in Example 4. The denominators tell you the step size.
8 · GCF/LCM mechanics, three ways.
9 · Same remainder, several divisors → LCM plus the remainder. "Leaves remainder $2$ when divided by $3,4,5,6$" means $N-2$ is a multiple of $\mathrm{lcm}(3,4,5,6)=60$, so $N=62$ (Practice 3). "Has $15$, $20$, and $25$ as factors" means a multiple of $\mathrm{lcm}=300$ (Practice 4). Board bonus: know the divisors of $60$ cold — $1,2,3,4,5,6,10,12,15,20,30,60$ — they unlock every minutes-per-mile and clock problem.

📖 Worked examples

2017 AMC 8 #9 · LCM of denominators

Example 1. All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles?

(A) 1(B) 2(C) 3(D) 4(E) 5
🧰 #7 LCM of denominators#1 Write the equation
Key idea: with $x$ marbles total and $y$ yellow, $\tfrac13x+\tfrac14x+6+y=x$, so $y=\tfrac{5}{12}x-6$. For $\tfrac13x$ and $\tfrac14x$ to be whole numbers, $x$ must be a multiple of $12$; and $y>0$ needs $\tfrac{5}{12}x>6$, i.e. $x>14.4$. The first multiple of $12$ past that is $x=24$, giving $y=\tfrac{5}{12}\cdot24-6=4$. Answer: D
2022 AMC 8 #13 · set up & count

Example 2. How many positive integers can fill the blank in the sentence below? "One positive integer is ___ more than twice another, and the sum of the two numbers is $28$."

(A) 6(B) 7(C) 8(D) 9(E) 10
🧰 #1 Write the equation#5 Count with bounds
Key idea: let the smaller number be $x$ and the blank be $y$: the other number is $2x+y$, and $(2x+y)+x=28$, i.e. $3x+y=28$. Each $x=1,2,\dots,9$ gives exactly one positive $y=28-3x$ ($x=9\Rightarrow y=1$; $x=10$ would force $y<0$). That's $9$ possible blanks. Answer: D
2015 AMC 8 #20 · system → one line

Example 3. Ralph went to the store and bought $12$ pairs of socks for a total of $\$24$. Some of the socks he bought cost $\$1$ a pair, some of the socks he bought cost $\$3$ a pair, and some of the socks he bought cost $\$4$ a pair. If he bought at least one pair of each type, how many pairs of $\$1$ socks did Ralph buy?

(A) 4(B) 5(C) 6(D) 7(E) 8
🧰 #6 Eliminate#2 Parity#5 Count with bounds
Key idea: with $x,y,z$ pairs at $\$1,\$3,\$4$: $x+y+z=12$ and $x+3y+4z=24$. Subtract: $2y+3z=12$. Parity: $3z$ must be even, so $z$ is even; $z\ge1$ forces $z=2$ ($z=4$ gives $y=0$, not allowed). Then $y=3$ and $x=12-3-2=7$. Answer: D
2019 AMC 8 #23 · bounds + integrality

Example 4. After Euclid High School's last basketball game, it was determined that $\frac14$ of the team's points were scored by Alexa and $\frac27$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $7$ team members?

(A) 10(B) 11(C) 12(D) 13(E) 14
🧰 #7 LCM of denominators#5 Count with bounds
Key idea: let the team total be $x$. The other seven scored $x-\tfrac14x-\tfrac27x-15=\tfrac{13}{28}x-15$, which must be an integer between $0$ and $14$. Integrality forces $28\mid x$ (steps of $28$!), and $0\le\tfrac{13}{28}x-15\le14$ pins $x=56$. Then the seven scored $\tfrac{13}{28}\cdot56-15=26-15=11$. Answer: B
2016 AMC 8 #20 · LCM constraints

Example 5. The least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$?

(A) 20(B) 30(C) 60(D) 120(E) 180
🧰 #8 GCF/LCM mechanics
Key idea: $\mathrm{lcm}(a,b)=2^2\cdot3$ and $\mathrm{lcm}(b,c)=3\cdot5$. The factor $2^2$ can only come from $a$ (since $b$ would push $2$ into $\mathrm{lcm}(b,c)$), and the $5$ can only come from $c$. So $4\mid a$ and $5\mid c$. Choose the leanest: $a=4$, $c=5$ (with $b=3$ making both LCMs work): $\mathrm{lcm}(a,c)=20$. Answer: A

✏️ Practice

Try each one before opening the solution. Figures are from the official exams.

2016 AMC 8 #11

1. Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132$.

(A) 5(B) 7(C) 9(D) 11(E) 12
Solution
🧰 #1 Write the equation#5 Count with bounds
Write the number as $10a+b$; reversed it is $10b+a$. Then $(10a+b)+(10b+a)=11(a+b)=132$, so $a+b=12$. Digits run $1$–$9$ here (with $a+b=12$ neither digit can be $0$), so the pairs are $(3,9),(4,8),(5,7),(6,6),(7,5),(8,4),(9,3)$ — $7$ numbers. Answer: B
2013 AMC 8 #10

2. What is the ratio of the least common multiple of $180$ and $594$ to the greatest common factor of $180$ and $594$?

(A) 110(B) 165(C) 330(D) 625(E) 660
Solution
🧰 #8 GCF/LCM mechanics
$180=2^2\cdot3^2\cdot5$ and $594=2\cdot3^3\cdot11$. LCM (larger exponents) $=2^2\cdot3^3\cdot5\cdot11$; GCF (smaller exponents) $=2\cdot3^2$. Ratio $=\dfrac{2^2\cdot3^3\cdot5\cdot11}{2\cdot3^2}=2\cdot3\cdot5\cdot11=330$. Answer: C
2012 AMC 8 #15

3. The smallest number greater than $2$ that leaves a remainder of $2$ when divided by $3$, $4$, $5$, or $6$ lies between what numbers?

(A) $40$ and $50$(B) $51$ and $55$(C) $56$ and $60$(D) $61$ and $65$(E) $66$ and $99$
Solution
🧰 #9 LCM + remainder
$N-2$ is divisible by $3,4,5,6$, so $N-2$ is a multiple of $\mathrm{lcm}(3,4,5,6)=60$. Smallest: $N=62$, which lies between $61$ and $65$. Answer: D
2003 AMC 8 #19

4. How many integers between $1000$ and $2000$ have all three of the numbers $15$, $20$, and $25$ as factors?

(A) 1(B) 2(C) 3(D) 4(E) 5
Solution
🧰 #9 LCM + remainder#8 GCF/LCM mechanics
Such a number is a multiple of $\mathrm{lcm}(15,20,25)$. With $15=3\cdot5$, $20=2^2\cdot5$, $25=5^2$: $\mathrm{lcm}=2^2\cdot3\cdot5^2=300$. Between $1000$ and $2000$: $1200$, $1500$, $1800$ — three of them. Answer: C
2022 AMC 8 #22

5. A bus takes $2$ minutes to drive from one stop to the next, and waits $1$ minute at each stop to let passengers board. Zia takes $5$ minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwise she will start walking toward the next stop. Suppose the bus and Zia start at the same time toward the library, with the bus $3$ stops behind. After how many minutes will Zia board the bus?

(A) 17(B) 19(C) 20(D) 21(E) 23
Solution
🧰 #9 LCM + remainder
The bus advances one stop every $3$ minutes (drive $2$, wait $1$); Zia advances one stop every $5$ minutes. So check the clock at $\mathrm{lcm}(3,5)=15$ minutes: the bus has made $5$ hops and Zia $3$ — starting $3$ stops back, the bus is now exactly one stop behind her, waiting to depart. By the rule Zia stays put, and the bus needs $2$ more minutes of driving to reach her: $15+2=17$. Answer: A

🆕 Fresh from the latest exams

The same ideas, exactly as they appeared on the most recent tests.

2025 AMC 8 #6

6. Sekou writes down the numbers $15, 16, 17, 18, 19$. After he erases one of his numbers, the sum of the remaining four numbers is a multiple of $4$. Which number did he erase?

(A) 15(B) 16(C) 17(D) 18(E) 19
Solution
🧰 #2 Parity / mod check
All five sum to $85$, and $85=84+1\equiv1\pmod4$. For the rest to be a multiple of $4$, the erased number must also be $\equiv1\pmod 4$ — that's $17$. Answer: C
2026 AMC 8 #18

7. 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
🧰 #1 Write the equation#3 Divisibility#5 Count with bounds
Let $k\ge2$ odds start at $a$: the sum is $ka+2(0+1+\cdots+(k-1))=k(a+k-1)=60$. So $k\mid 60$ and $a=\tfrac{60}{k}-k+1$ must be a positive odd integer. Run the divisors: $k=2\Rightarrow a=29$ ✓ ($29+31$); $k=3\Rightarrow18$ ✗; $k=4\Rightarrow12$ ✗; $k=5\Rightarrow8$ ✗; $k=6\Rightarrow5$ ✓ ($5+7+9+11+13+15$); $k=10\Rightarrow a<0$. Two ways. Answer: B
2024 AMC 8 #15

8. Let the letters $F,L,Y,B,U,G$ represent distinct digits. Suppose $\underline{F}\,\underline{L}\,\underline{Y}\,\underline{F}\,\underline{L}\,\underline{Y}$ is the GREATEST number that satisfies the equation \[8\cdot \underline{F}\,\underline{L}\,\underline{Y}\,\underline{F}\,\underline{L}\,\underline{Y}=\underline{B}\,\underline{U}\,\underline{G}\,\underline{B}\,\underline{U}\,\underline{G}.\] What is the value of $\underline{F}\,\underline{L}\,\underline{Y}+\underline{B}\,\underline{U}\,\underline{G}$?

(A) 1089(B) 1098(C) 1107(D) 1116(E) 1125
Solution
🧰 #8 Factor structure#5 Count with bounds
The repeat pattern is a factorization in disguise: $\underline{FLYFLY}=1001\cdot\underline{FLY}$ and $\underline{BUGBUG}=1001\cdot\underline{BUG}$. Cancel $1001$: the equation is just $8\cdot\underline{FLY}=\underline{BUG}$. For $\underline{BUG}$ to stay three digits, $\underline{FLY}\le124$. Test downward with distinct digits: $124\to992$ (repeated $9$) ✗; $123\to984$ — digits $1,2,3,9,8,4$ all distinct ✓. $123+984=1107$. Answer: C

⭐ Challenge (AMC 10 level)

2017 AMC 10A #16 · LCM

C1. There are $10$ horses, named Horse $1$, Horse $2,\dots,$ Horse $10$. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point. The horses run in the same direction at their constant speeds. The least time $S>0$, in minutes, at which all $10$ horses are again simultaneously at the starting point is $S=2520$. Let $T>0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T$?

(A) 2(B) 3(C) 4(D) 5(E) 6
Solution
🧰 #9 LCM + remainder#8 GCF/LCM mechanics
Horse $k$ is at the start exactly at the multiples of $k$, so $T$ works iff at least $5$ of the numbers $1,\dots,10$ divide $T$ — i.e. $T$ has $5$ divisors that are $\le 10$. Hunt for the smallest such $T$: $12$ is divisible by $1,2,3,4,6$ — five horses — and no smaller number has five divisors. $T=12$, digit sum $1+2=3$. Answer: B

🗝 Answer key

Example12345
DDDBA
Practice12345678
BCDCACBC
ChallengeC1
B

Sources: prep-course Lecture 16 (video + board notes + 11 worked examples, translated to English) and official AMC 8 / AMC 10 problems from the local bank (kb/content/banks/). Answers verified against the bank's official answer keys where available.