Worked examples — Modular arithmetic — definition, addition, multiplication, congruence

This page is a drill floor. The parent note taught you the rules; here we throw every kind of input at those rules — positive, negative, zero, huge powers, word problems, and one exam trap — so that no future problem surprises you.

Before any symbol appears, one reminder in plain words: writing $a\equiv b(\mathrm{mod}n)$ means "$a$ and $b$ land on the same spot on a circle with $n$ marks numbered $0,1,2,\dots ,n-1$." The mark you land on is the remainder after dividing by $n$. That circle is the picture behind every example below.

Look at the red mark: it is position $3$ on a circle of $n=7$ marks. Every number that leaves remainder $3$ when divided by $7$ — that is $3,10,17,24,\dots$ and also $-4,-11,\dots$ — sits on that same red mark. That is the whole idea of a "bucket."

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The scenario matrix

Every problem this topic can throw is one of these cells. Each worked example below is tagged with the cell it covers.

Cell	Scenario	Danger it tests
A	Positive $+$ positive	reduce-first habit
B	Positive $\times$ positive	reduce both factors
C	Negative input	wrapping into $[0,n-1]$
D	Subtraction (a special negative)	$a-b$ can go below zero
E	Large power, base $\equiv 1$	trivial-base shortcut
F	Large power, base $\not\equiv \pm 1$	finding the cycle length
G	Zero / degenerate ($a\equiv 0$, or $n=1$)	edge behaviour
H	Word problem (real-world clock/calendar)	translating to $\mathrm{mod}$
I	Exam twist — illegal "division"	the coprime condition

Nine cells. Nine (and one bonus) worked examples follow. Prerequisites you may want open: Division Algorithm, GCD and LCM, Modular Inverses.

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Example 1 — Cell A (positive + positive)

Forecast: Guess the final answer (a number between $0$ and $8$) before reading on.

1. Reduce each operand. $48=5\times 9+3$, so $48\equiv 3$. And $65=7\times 9+2$, so $65\equiv 2(\mathrm{mod}9)$. Why this step? The addition rule says congruent numbers may be swapped for their remainders before adding — smaller numbers are safer to handle.
2. Add the small numbers. $3+2=5$. Why this step? By the addition rule $(48+65)\equiv (3+2)(\mathrm{mod}9)$.
3. Is it in range? $5<9$, so we stop. Why this step? Standard form lives in $[0,8]$.

Answer: $(48+65)\equiv 5(\mathrm{mod}9)$.

Verify: $48+65=113=12\times 9+5$. Remainder $5$. ✓

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Example 2 — Cell B (positive × positive)

Forecast: Multiplying gives $884$ — but you should be able to name the remainder without that.

1. Reduce each factor. $26=2\times 11+4\Rightarrow 26\equiv 4$. $34=3\times 11+1\Rightarrow 34\equiv 1(\mathrm{mod}11)$. Why this step? The multiplication rule lets us replace each factor by its remainder.
2. Multiply the reduced values. $4\times 1=4$. Why this step? $(26\times 34)\equiv (4\times 1)(\mathrm{mod}11)$.
3. Reduce the product. $4<11$, already standard.

Answer: $(26\times 34)\equiv 4(\mathrm{mod}11)$.

Verify: $26\times 34=884=80\times 11+4$. Remainder $4$. ✓

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Example 3 — Cell C (negative input)

Forecast: A negative number must still land on a mark in $\{0,1,2,3,4,5\}$. Which one?

1. Divide, allowing a negative quotient. We need $-47=q\times 6+r$ with $0\le r<6$. Try $q=-8$: $-8\times 6=-48$, and $-47-(-48)=1$. Why this step? The Division Algorithm guarantees a unique remainder in $[0,5]$ even for negatives — we just pick $q$ small enough.
2. Read off the remainder. $r=1$, so $-47\equiv 1(\mathrm{mod}6)$. Why this step? That $r$ is the mark we land on when stepping backwards around the circle (red arrow in the figure).
3. Shortcut check. Equivalently, add multiples of $6$ until non-negative: $-47+8\times 6=-47+48=1$. Same answer.

Answer: $-47\equiv 1(\mathrm{mod}6)$.

Verify: $1-(-47)=48=8\times 6$, a multiple of $6$. ✓

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Example 4 — Cell D (subtraction dipping below zero)

Forecast: Subtraction is just "add a negative," so expect a wrap-around.

1. Reduce each term. $9\equiv 9$ (already small); $20=1\times 13+7\Rightarrow 20\equiv 7(\mathrm{mod}13)$. Why this step? Reduce before operating, exactly as with addition — subtraction obeys the same swap rule.
2. Subtract the reduced values. $9-7=2$. Why this step? $(9-20)\equiv (9-7)(\mathrm{mod}13)$.
3. Fix the sign if needed. Here $2\ge 0$, so no wrap is required.

Answer: $(9-20)\equiv 2(\mathrm{mod}13)$.

Verify: $9-20=-11$, and $-11+13=2$. So $-11\equiv 2$. ✓

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Example 5 — Cell E (large power, base ≡ 1)

Forecast: An exponent of $500$ looks terrifying. It won't be.

1. Reduce the base. $8=1\times 7+1\Rightarrow 8\equiv 1(\mathrm{mod}7)$. Why this step? If the base is $\equiv 1$, the power rule collapses everything.
2. Apply the power rule. $8^{500}\equiv 1^{500}(\mathrm{mod}7)$. Why this step? Congruence survives exponentiation: $a\equiv b\Rightarrow a^k\equiv b^k$.
3. Evaluate. $1^{500}=1$.

Answer: $8^{500}\equiv 1(\mathrm{mod}7)$.

Verify: Since $8\equiv 1$, every power of $8$ is $\equiv 1$. Check tiny case $8^2=64=9\times 7+1\equiv 1$. ✓ (Compare with Fermats Little Theorem for a deeper reason powers cycle.)

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Example 6 — Cell F (large power, base ≢ ±1 — find the cycle)

Forecast: Here $3\not\equiv \pm 1(\mathrm{mod}7)$, so we cannot collapse the base. Guess whether the answer is small.

1. List the powers of $3$ until they repeat. $3^1\equiv 3$, $3^2\equiv 9\equiv 2$, $3^3\equiv 6$, $3^4\equiv 18\equiv 4$, $3^5\equiv 12\equiv 5$, $3^6\equiv 15\equiv 1(\mathrm{mod}7)$. Why this step? Powers must cycle (only $7$ possible remainders), and the moment we hit $1$ the pattern restarts. The cycle length here is $6$.
2. Use the cycle to shrink the exponent. Since $3^6\equiv 1$, reduce $100$ modulo the cycle length $6$: $100=16\times 6+4$, so the leftover exponent is $4$. Why this step? $3^{100}=(3^6{)}^{16}\cdot 3^4\equiv 1^{16}\cdot 3^4\equiv 3^4(\mathrm{mod}7)$.
3. Read the answer off the list. From step 1, $3^4\equiv 4$.

Answer: $3^{100}\equiv 4(\mathrm{mod}7)$.

Verify: The cycle $3,2,6,4,5,1$ (red loop in the figure) repeats every $6$; position $100\mathrm{mod}6=4$ gives the $4$th entry $=4$. ✓

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Example 7 — Cell G (zero / degenerate)

Forecast: Edge cases. Do the rules still hold when a factor is $0$, or the modulus is $1$?

1. Part (a): a zero factor. $0\equiv 0$, so $0\times 55\equiv 0\times (55\mathrm{mod}9)\equiv 0(\mathrm{mod}9)$. Why this step? The multiplication rule works for any residues, including $0$. Zero times anything is $0$, and $0$ is already standard.
2. Part (b): modulus $1$. With $n=1$, the marks on the circle are just $\{0\}$ — a single point. Every integer leaves remainder $0$. Why this step? $17=17\times 1+0$, so $17\equiv 0(\mathrm{mod}1)$. In mod $1$ arithmetic all numbers collapse to one class.

Answers: (a) $0$. (b) $0$.

Verify: (a) $0\times 55=0=0\times 9+0$. ✓ (b) $17-0=17=17\times 1$, a multiple of $1$. ✓

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Example 8 — Cell H (real-world word problem)

Forecast: Weeks are a circle of $7$ marks. Guess a day.

1. Model days as marks mod 7. Number the days $0=$ Wed, $1=$ Thu, $2=$ Fri, $3=$ Sat, $4=$ Sun, $5=$ Mon, $6=$ Tue. Why this step? A week is a $7$-mark clock; "in $1000$ days" means stepping $1000$ marks forward.
2. Reduce the step count. $1000=142\times 7+6$, so $1000\equiv 6(\mathrm{mod}7)$. Why this step? Whole weeks ($142$ of them) return you to the same day, so only the leftover $6$ steps matter.
3. Step forward from Wed. $0+6=6=$ Tuesday.

Answer: In $1000$ days it will be Tuesday.

Verify: $1000=142\times 7+6$ and $142\times 7=994$, $994+6=1000$. ✓ Six days after Wednesday: Thu, Fri, Sat, Sun, Mon, Tue. ✓

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Example 9 — Cell I (exam twist: illegal division)

Forecast: Your instinct says "divide both sides by $6$." Resist — check the gcd first.

1. Check whether you may divide by $6$. $\gcd (6,15)=3\ne 1$, so $6$ has no modular inverse mod $15$; naive division is forbidden. Why this step? Cancelling by $a$ is only legal when $\gcd (a,n)=1$ (see GCD and LCM).
2. Divide the whole congruence by the common factor $3$. Since $3$ divides $6$, $9$, and $15$: $2x\equiv 3(\mathrm{mod}5)$. Why this step? When $d=\gcd (a,n)$ divides $b$, you may divide $a$, $b$, and $n$ all by $d$. This is the legal move — a Linear Congruences technique.
3. Now $\gcd (2,5)=1$, so invert $2$ mod $5$. We need $2y\equiv 1(\mathrm{mod}5)$; testing, $2\times 3=6\equiv 1$, so $2^{-1}\equiv 3$. Why this step? With a genuine inverse we may finally "divide."
4. Multiply through by $3$. $x\equiv 3\times 3=9\equiv 4(\mathrm{mod}5)$.

Answer: $x\equiv 4(\mathrm{mod}5)$ — equivalently $x\in \{4,9,14\}$ as residues mod $15$.

Verify: Take $x=4$: $6\times 4=24=15+9\equiv 9(\mathrm{mod}15)$. ✓ Take $x=9$: $6\times 9=54=3\times 15+9\equiv 9$. ✓ Take $x=14$: $6\times 14=84=5\times 15+9\equiv 9$. ✓

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Recall Self-test (reveal after answering)

How do you reduce a negative number to standard form? ::: Add the modulus repeatedly until the result lands in $[0,n-1]$. Why can $8^{500}\equiv 1(\mathrm{mod}7)$ be found instantly? ::: Because $8\equiv 1$, and $1$ to any power is $1$. For $3^{100}\mathrm{mod}7$, what number do you reduce the exponent $100$ by? ::: The cycle length $6$ (since $3^6\equiv 1$), giving leftover exponent $4$. When is dividing both sides of a congruence by $a$ legal? ::: Only when $\gcd (a,n)=1$.