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Remainders & Divisibility · Lecture 17 · nt-remainders-divisibility

📺 Course materials — Lecture 17: lecture video · handout PDF (Lecture 17) · board notes (17.jpg) — all in the amc8/video folder. (This lecture has no homework-solutions chapter; the examples below were rebuilt from the board notes and the official problem bank.)  ·  🖨 Printable PDF (answers & solutions at the back)

🔑 Key facts

Every division fits one template: $\;N = d\cdot q + r$ with $0 \le r < d$ — dividend $=$ divisor $\times$ quotient $+$ remainder, and the remainder is always smaller than the divisor. "$d$ divides $N$" just means $r = 0$. Two workhorses fall straight out of the template: the units digit of $N$ is its remainder mod $10$, and remainders repeat in a cycle of length $d$ — so almost every remainder problem is really a problem about finding the cycle.

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

1 · Anatomy of a remainder. $20 \div 6 = 3$ r $2$ means $20 = 6\times3 + 2$ — and the instructor's first warning: $20 = 6\times2 + 8$ is not a legal answer, because a remainder must be smaller than the divisor. When a problem gives you a quotient-and-remainder fact, immediately rewrite it as $N = dq + r$; the equation form is what you can compute with.
2 · How remainders scale. From the board, with $A \div B = C$ r $D$: The trap the instructor flags in red: scaling the divisor alone wrecks both quotient and remainder — no shortcut there.
3 · The three divisibility-rule families (the board's table — knowing why tells you when): For a composite divisor, split it into coprime parts and test both: $6 = 2\cdot3$, $15 = 3\cdot5$, $12 = 3\cdot4$ (Example 3).
4 · The chop-and-double rule for 7. Cut off the units digit and subtract it twice from what's left; repeat. $3185 \to 318 - 2\times5 = 308 \to 30 - 2\times8 = 14$ — a multiple of $7$, so $3185$ is too. It works because $\overline{ab} \;(= 10a + b)$ and $a - 2b$ differ by $21b$, a multiple of $7$.
5 · Units digits of powers run in short cycles. Only the units digit of the base matters ($2023^7$ ends like $3^7$), and every digit's cycle has length $1$, $2$ or $4$. So reduce the base, find the small cycle, then divide the exponent by the cycle length and read the remainder: $13^{2012}$ ends like $3^{2012}$, cycle $3,9,7,1$, and $2012$ is a multiple of $4$ — units digit $1$. Likewise $2024^{2024}$ ends like $4^{2024}$, cycle $4,6$ — units digit $6$.
3971 powers of 3: cycle length 4
6 · Same remainder (or same deficit) → jump to the LCM. If $N$ leaves the same remainder $r$ when divided by several numbers, then $N - r$ is a multiple of their LCM (Practice 2: remainder $1$ mod $4,5,6$ ⟹ $N-1$ is a multiple of $60$). If instead $N$ falls short by the same amount $k$ (remainder $=$ divisor $-\,k$ each time), then $N + k$ is a multiple of the LCM (Practice 9: remainders $2, 5, 7$ for $6, 9, 11$ are all "$4$ short" ⟹ $198 \mid N+4$). Build the LCM from prime powers, never by multiplying everything together.
7 · Repeat-block numbers factor instantly. $\overline{abcabc} = \overline{abc}\times1001 = \overline{abc}\times7\times11\times13$ — memorize $1001 = 7\times11\times13$. Same idea one size down: $\overline{abab} = \overline{ab}\times101$. Any "copy-paste" number is a small number times a known constant (Example 2).
8 · Count multiples by first and last index. How many multiples of $13$ are three-digit? The first is $13\times8 = 104$, the last is $13\times76 = 988$ (check: $13\times77 = 1001$, too big), so there are $76 - 8 + 1 = 69$. The count formula $\lfloor N/k \rfloor$ counts multiples of $k$ up to $N$; the index trick avoids off-by-one errors in a range (Practice 5).
9 · Weekdays are mod 7. Calendar problems are remainder problems in disguise: step forward $10$ days $=$ step forward $3$ weekdays. List each event's offset mod $7$ and compare against the forbidden (or required) day — that's the whole of Practice 8.
10 · Count a prime inside $n!$ with floors (Legendre). The exponent of $p$ in $n!$ is \[\left\lfloor \tfrac{n}{p}\right\rfloor + \left\lfloor \tfrac{n}{p^2}\right\rfloor + \left\lfloor \tfrac{n}{p^3}\right\rfloor + \cdots\] Board examples: fives in $98!$: $\lfloor 98/5\rfloor + \lfloor 98/25\rfloor = 19 + 3 = 22$; sevens in $2024!$: $289 + 41 + 5 = 335$. Each floor counts one "layer": multiples of $p$, then the extra factor in multiples of $p^2$, and so on (Example 4).

📖 Worked examples

2016 AMC 8 #5 · translate to digits

Example 1. The number $N$ is a two-digit number. When $N$ is divided by $9$, the remainder is $1$. When $N$ is divided by $10$, the remainder is $3$. What is the remainder when $N$ is divided by $11$?

(A) 0(B) 2(C) 4(D) 5(E) 7
🧰 #1 Anatomy#3 Rule families
Key idea: each remainder fact is a digit fact. "Remainder $3$ mod $10$" says the units digit is $3$; "remainder $1$ mod $9$" says the digit sum is $1$ more than a multiple of $9$. Scan $13, 23, 33, \ldots$: only $73$ has digit sum $10$ ✓. Then $73 = 11\times6 + 7$. Answer: E
2017 AMC 8 #7 · repeat-block

Example 2. Let $Z$ be a $6$-digit positive integer, such as $247247$, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$?

(A) 11(B) 19(C) 101(D) 111(E) 1111
🧰 #7 Repeat-block
Key idea: copy-pasting a $3$-digit block is multiplication: \[Z = \overline{abc}\times1000 + \overline{abc} = \overline{abc}\times1001 = \overline{abc}\times7\times11\times13.\] So $7$, $11$ and $13$ all divide every such $Z$ — and $11$ is on the list. (The others aren't guaranteed: $247247/101$ isn't even an integer.) Answer: A
2020 AMC 8 #19 · split the divisor

Example 3. A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15$?

(A) 3(B) 4(C) 5(D) 6(E) 8
🧰 #3 Rule families
Key idea: $15 = 3\times5$ — test both parts. A five-digit flippy number looks like $\overline{ababa}$. Divisible by $5$ ⟹ it ends in $0$ or $5$; the last digit is $a$, which is also the leading digit, so $a \ne 0$, forcing $a = 5$: the number is $\overline{5b5b5}$. Divisible by $3$ ⟹ digit sum $15 + 2b$ is a multiple of $3$ ⟹ $b \in \{0, 3, 6, 9\}$ — four numbers ($50505, 53535, 56565, 59595$). Answer: B
2017 AMC 8 #19 · factorials

Example 4. For any positive integer $M$, the notation $M!$ denotes the product of the integers $1$ through $M$. What is the largest integer $n$ for which $5^n$ is a factor of the sum $98! + 99! + 100!$?

(A) 23(B) 24(C) 25(D) 26(E) 27
🧰 #10 Legendre
Key idea: factor out the smallest factorial first: \[98! + 99! + 100! = 98!\,(1 + 99 + 99\times100) = 98!\times10000 = 98!\times2^4\times5^4.\] Now count the fives inside $98!$ with floors: $\lfloor 98/5\rfloor + \lfloor 98/25\rfloor = 19 + 3 = 22$. Add the $5^4$ from the $10000$: $\;22 + 4 = 26$. Answer: D

✏️ Practice

Try each one before opening the solution.

2016 AMC 8 #20

1. 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
Solution
🧰 #6 Build the LCM
$b$ divides both $12 = 2^2\cdot3$ and $15 = 3\cdot5$, so $b$ is $1$ or $3$ — $b$ carries no $2$s and no $5$s. Then $a$ must supply the whole $2^2$ (so $4 \mid a$) and $c$ must supply the $5$ (so $5 \mid c$). Choosing $a = 4$, $b = 3$, $c = 5$ works, and $\operatorname{lcm}(4,5) = 20$. Answer: A
2017 AMC 8 #12

2. The smallest positive integer greater than $1$ that leaves a remainder of $1$ when divided by $4$, $5$, and $6$ lies between which of the following pairs of numbers?

(A) $2$ and $19$(B) $20$ and $39$(C) $40$ and $59$(D) $60$ and $79$(E) $80$ and $124$
Solution
🧰 #6 LCM shift
Same remainder $1$ for all three divisors ⟹ $N - 1$ is a common multiple of $4, 5, 6$. The smallest positive one is $\operatorname{lcm}(4,5,6) = 2^2\cdot3\cdot5 = 60$ (not $4\cdot5\cdot6 = 120$ — don't double-count the shared $2$!). So $N = 61$, which lies between $60$ and $79$. Answer: D
2018 AMC 8 #7

3. The $5$-digit number $\underline{2}\,\underline{0}\,\underline{1}\,\underline{8}\,\underline{U}$ is divisible by $9$. What is the remainder when this number is divided by $8$?

(A) 1(B) 3(C) 5(D) 6(E) 7
Solution
🧰 #3 Rule families
Digit-sum rule: $2+0+1+8+U = 11+U$ must be a multiple of $9$, so $U = 7$. Last-three-digits rule for $8$: $187 = 8\times23 + 3$ — remainder $3$. Answer: B
mock · course problem

4. What is the units digit of $13^{2012}$?

(A) 1(B) 3(C) 5(D) 7(E) 9
Solution
🧰 #5 Power cycles
Only the base's units digit matters: $13^{2012}$ ends like $3^{2012}$. Powers of $3$ end $3, 9, 7, 1, 3, 9, \ldots$ — cycle length $4$. Since $2012$ is a multiple of $4$, we land on the last digit of the cycle: $1$. Answer: A
mock · course problem

5. How many three-digit numbers are divisible by $13$?

(A) 7(B) 67(C) 69(D) 76(E) 77
Solution
🧰 #8 First & last index
First three-digit multiple: $13\times8 = 104$. Last: $13\times76 = 988$ (since $13\times77 = 1001$). Count the indices: $76 - 8 + 1 = 69$. Answer: C
2022 AMC 8 #17

6. If $n$ is an even positive integer, the double factorial notation $n!!$ represents the product of all the even integers from $2$ to $n$. For example, $8!! = 2 \cdot 4 \cdot 6 \cdot 8$. What is the units digit of the following sum? \[2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!\]

(A) 0(B) 2(C) 4(D) 6(E) 8
Solution
🧰 #5 Units digits
From $10!!$ onward every term contains the factor $10$, so those units digits are all $0$. Only four terms matter: $2!! = 2$, $4!! = 8$, $6!! = 48$, $8!! = 384$ — units digits $2 + 8 + 8 + 4 = 22$, which ends in $2$. Answer: B
2016 AMC 8 #24

7. The digits $1$, $2$, $3$, $4$, and $5$ are each used once to write a five-digit number $PQRST$. The three-digit number $PQR$ is divisible by $4$, the three-digit number $QRS$ is divisible by $5$, and the three-digit number $RST$ is divisible by $3$. What is $P$?

(A) 1(B) 2(C) 3(D) 4(E) 5
Solution
🧰 #3 Rule families
Divisible by $5$ ⟹ $S = 5$ (no digit $0$ available). Divisible by $4$ ⟹ the two-digit tail $\overline{QR}$ is a multiple of $4$: from the digits $1$–$4$ that's $12$, $24$ or $32$. Divisible by $3$ ⟹ $R + S + T = R + 5 + T$ is a multiple of $3$. Test: $QR = 12$ leaves $T \in \{3,4\}$, sums $10, 11$ ✗; $QR = 32$ leaves $T\in\{1,4\}$, sums $8, 11$ ✗; $QR = 24$ leaves $T = 3$: $4+5+3 = 12$ ✓. So $Q=2, R=4, T=3$ and $P$ is the leftover digit: $1$. Answer: A
2019 AMC 8 #14

8. Isabella has $6$ coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every $10$ days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the $6$ dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?

(A) Monday(B) Tuesday(C) Wednesday(D) Thursday(E) Friday
Solution
🧰 #9 Mod 7 calendar
The six dates are $0, 10, 20, 30, 40, 50$ days after the first — mod $7$ that's $0, 3, 6, 2, 5, 1$: six different weekday shifts, missing only $4$. For no date to be a Sunday, Sunday must sit exactly $4$ weekdays after the first coupon's day — and $4$ days before Sunday is Wednesday. Answer: C
2018 AMC 8 #21

9. How many positive three-digit integers have a remainder of $2$ when divided by $6$, a remainder of $5$ when divided by $9$, and a remainder of $7$ when divided by $11$?

(A) 1(B) 2(C) 3(D) 4(E) 5
Solution
🧰 #6 Same deficit
Spot the pattern: $2 = 6-4$, $\;5 = 9-4$, $\;7 = 11-4$ — every division falls short by the same $4$. So $N + 4$ is divisible by $6$, $9$ and $11$, i.e. by $\operatorname{lcm}(6,9,11) = 198$. Three-digit candidates: $N + 4 = 198, 396, 594, 792, 990$ — five values ($N = 194, \ldots, 986$). Answer: E
2020 AMC 8 #12

10. For a positive integer $n$, the factorial notation $n!$ represents the product of the integers from $n$ to $1$. (For example, $6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$.) What value of $N$ satisfies the following equation? \[5! \cdot 9! = 12 \cdot N!\]

(A) 10(B) 11(C) 12(D) 13(E) 14
Solution
🧰 #10 Factorials
$5! = 120 = 12\times10$, so $5!\cdot9! = 12\times10\times9! = 12\times10!$ — matching $12\cdot N!$ gives $N = 10$. Answer: A

🆕 Fresh from the latest exams

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

2024 AMC 8 #1

11. What is the ones digit of \[222{,}222-22{,}222-2{,}222-222-22-2\,?\]

(A) 0(B) 2(C) 4(D) 6(E) 8
Solution
🧰 #5 Units digits
Group the subtractions: $22222 + 2222 + 222 + 22 + 2 = 24690$, and $222222 - 24690 = 197532$ — ones digit $2$. (Units-only shortcut: $2 - (2+2+2+2+2) = 2 - 10 \equiv 2 \pmod{10}$, watching the borrows.) Answer: B
2025 AMC 8 #6

12. 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
🧰 #1 Anatomy#6 LCM shift
The five numbers add to $85$. Erasing $x$ must leave a multiple of $4$: $85 - x \equiv 0 \pmod 4$, and $85 = 4\times21 + 1$, so $x \equiv 1 \pmod 4$ — among the choices only $17$. Check: $85 - 17 = 68 = 4\times17$ ✓. Answer: C
2026 AMC 8 #16

13. Consider all positive four-digit integers consisting of only even digits. What fraction of these integers are divisible by $4$?

(A) $\frac{1}{4}$(B) $\frac{2}{5}$(C) $\frac{1}{2}$(D) $\frac{3}{5}$(E) $\frac{3}{4}$
Solution
🧰 #3 Rule families
Divisibility by $4$ only looks at the last two digits: $\overline{tu} = 10t + u \equiv 2t + u \pmod 4$. Here the tens digit $t$ is even, so $2t$ is already a multiple of $4$ — the condition collapses to $4 \mid u$, i.e. $u \in \{0, 4, 8\}$: $3$ of the $5$ possible even units digits. The first three digits never matter, so the fraction is $\tfrac35$. Answer: D

🗝 Answer key

Example1234
EABD
Practice12345678910111213
ADBACBACEABCD

Sources: prep-course Lecture 17 (video + board notes, translated to English — this lecture has no homework-solutions chapter, so the worked examples were rebuilt from the board notes) and official AMC 8 problems from the local bank (kb/content/banks/). Answers verified against the bank's official answer keys. Companion handouts: Counting Divisors and Counting Multiples & Divisibility.