Congruences

This chapter introduces the fundamental concept of congruences, covering notation, modular operations, and linear congruences. A thorough understanding of these principles is crucial for advanced number theory topics and is frequently assessed in CMI examinations.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Congruence notation | | 2 | Operations modulo n | | 3 | Remainder patterns | | 4 | Linear congruences |

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We begin with Congruence notation.

Part 1: Congruence notation

Congruence Notation

Overview

Congruence notation is the language used to express equality of remainders. It is one of the central ideas of modular arithmetic. In exam problems, correct use of notation matters: writing $\qquad a \equiv b \pmod n$ does not mean $a=b$; it means $a$ and $b$ differ by a multiple of $n$. ---

Learning Objectives

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Definition

That is, $a-b$ is divisible by $n$. ::: Examples:

• $\qquad 17 \equiv 5 \pmod{12}$ because $17-5=12$

• $\qquad -1 \equiv 4 \pmod 5$ because $-1-4=-5$

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Congruence Is Not Equality

For example, $\qquad 14 \equiv 2 \pmod{12}$ but clearly $\qquad 14 \ne 2$ ::: ---

Equivalent Forms

These are three different ways of saying the same thing. ::: ---

Basic Properties

So congruence modulo $n$ is an equivalence relation on the integers. ::: ---

Compatibility with Operations

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Least Residue Language

This is why we often work with the standard set $\qquad \{0,1,2,\dots,n-1\}$ ::: ---

Minimal Worked Examples

Example 1 Show that $\qquad 38 \equiv 3 \pmod 5$ We check: $\qquad 38-3=35$ and $35$ is divisible by $5$. So $\qquad 38 \equiv 3 \pmod 5$ --- Example 2 Show that $\qquad -17 \equiv 3 \pmod 5$ We compute: $\qquad -17-3=-20$ which is divisible by $5$. Hence $\qquad -17 \equiv 3 \pmod 5$ ---

CMI Strategy

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="The statement $17\equiv 5 \pmod{12}$ means" options=["$17=5$","$12\mid (17-5)$","$12\mid (17+5)$","$17$ and $5$ are both divisible by $12$"] answer="B" hint="Use the definition of congruence." solution="By definition, $\qquad 17 \equiv 5 \pmod{12}$ means $\qquad 12 \mid (17-5)$ Since $17-5=12$, this is true. Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="What is the greatest positive integer $n$ such that $38\equiv 3 \pmod n$?" answer="35" hint="Use the divisibility form." solution="The statement $\qquad 38 \equiv 3 \pmod n$ means $\qquad n \mid (38-3)=35$ So $n$ must be a positive divisor of $35$. The greatest such $n$ is $\boxed{35}$." ::: :::question type="MSQ" question="Which of the following are always true?" options=["If $a\equiv b \pmod n$, then $n\mid (a-b)$","If $a\equiv b \pmod n$, then $a$ and $b$ leave the same remainder upon division by $n$","If $a\equiv b \pmod n$, then $a=b$","If $a\equiv b \pmod n$ and $b\equiv c \pmod n$, then $a\equiv c \pmod n$"] answer="A,B,D" hint="Check definition and transitivity." solution="1. True by definition.

True.

False. Congruence is weaker than equality.

True by transitivity.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Prove that if $a\equiv b \pmod n$, then $a^2\equiv b^2 \pmod n$." answer="Since $a-b$ is divisible by $n$, so is $(a-b)(a+b)=a^2-b^2$." hint="Factor $a^2-b^2$." solution="Assume $\qquad a \equiv b \pmod n$ Then by definition, $\qquad n \mid (a-b)$ Now $\qquad a^2-b^2=(a-b)(a+b)$ Since $n$ divides $a-b$, it must also divide $(a-b)(a+b)$. Therefore $\qquad n \mid (a^2-b^2)$ Hence $\qquad a^2 \equiv b^2 \pmod n$ as required." ::: ---

Summary

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Part 2: Operations modulo n

Operations Modulo $n$

Overview

Working modulo $n$ means we care only about remainders after division by $n$. The real power of modular arithmetic is that addition, subtraction, and multiplication can all be performed on remainders instead of large numbers. In exam problems, this is one of the fastest ways to simplify computations, detect impossibility, and reduce long expressions. ---

Learning Objectives

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Core Idea

If $\qquad a \equiv r \pmod n$, then in modular calculations we may replace $a$ by $r$. ::: ---

Addition, Subtraction, Multiplication

This means we can reduce numbers before operating. ---

Reduction Rule

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Powers Modulo $n$

Example: Since $\qquad 17 \equiv 2 \pmod 5$, we get $\qquad 17^3 \equiv 2^3 = 8 \equiv 3 \pmod 5$ ::: ---

Least Nonnegative Residue

Examples:

• modulo $5$, the least nonnegative residue of $17$ is $2$

• modulo $7$, the least nonnegative residue of $-3$ is $4$

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Cancellation: Be Careful

Example: Modulo $6$, $\qquad 2\cdot 1 \equiv 2\cdot 4 \pmod 6$ because both sides are congruent to $2$ mod $6$. But $\qquad 1 \not\equiv 4 \pmod 6$ So cancellation failed. ---

Minimal Worked Examples

Example 1 Compute $\qquad 23+19 \pmod 6$ Reduce first: $\qquad 23 \equiv 5 \pmod 6,\qquad 19 \equiv 1 \pmod 6$ So $\qquad 23+19 \equiv 5+1 = 6 \equiv 0 \pmod 6$ Hence the answer is $\boxed{0}$. --- Example 2 Compute $\qquad 14\cdot 17 \pmod 5$ Reduce first: $\qquad 14 \equiv 4 \pmod 5,\qquad 17 \equiv 2 \pmod 5$ So $\qquad 14\cdot 17 \equiv 4\cdot 2 = 8 \equiv 3 \pmod 5$ Hence the answer is $\boxed{3}$. ---

CMI Strategy

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="The least nonnegative residue of $29+17$ modulo $6$ is" options=["$1$","$2$","$4$","$5$"] answer="C" hint="Reduce before adding." solution="We have $\qquad 29 \equiv 5 \pmod 6,\qquad 17 \equiv 5 \pmod 6$ So $\qquad 29+17 \equiv 5+5 = 10 \equiv 4 \pmod 6$ Hence the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Find the least nonnegative residue of $17\cdot 23+5$ modulo $6$." answer="0" hint="Reduce the factors modulo $6$ first." solution="Reduce first: $\qquad 17 \equiv 5 \pmod 6,\qquad 23 \equiv 5 \pmod 6$ So $\qquad 17\cdot 23+5 \equiv 5\cdot 5+5 = 25+5 = 30 \equiv 0 \pmod 6$ Hence the required residue is $\boxed{0}$." ::: :::question type="MSQ" question="Which of the following are always valid modulo $n$?" options=["If $a\equiv b \pmod n$ and $c\equiv d \pmod n$, then $a+c\equiv b+d \pmod n$","If $a\equiv b \pmod n$ and $c\equiv d \pmod n$, then $ac\equiv bd \pmod n$","From $ac\equiv bc \pmod n$, we may always cancel $c$","If $a\equiv b \pmod n$, then $a^2\equiv b^2 \pmod n$"] answer="A,B,D" hint="Check which operations are universally compatible." solution="1. True.

True.

False. Cancellation is not always valid.

True, because squaring preserves congruence.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Find the least nonnegative residue of $23^2+7\cdot 11$ modulo $8$." answer="$6$" hint="Reduce each part modulo $8$ first." solution="Reduce first: $\qquad 23 \equiv 7 \pmod 8$ So $\qquad 23^2 \equiv 7^2 = 49 \equiv 1 \pmod 8$ Also $\qquad 7\cdot 11 \equiv 7\cdot 3 = 21 \equiv 5 \pmod 8$ Therefore $\qquad 23^2+7\cdot 11 \equiv 1+5 = 6 \pmod 8$ So the least nonnegative residue is $\boxed{6}$." ::: ---

Summary

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Part 3: Remainder patterns

Remainder Patterns

Overview

Remainder-pattern problems ask you to understand how a sequence behaves modulo some integer. At exam level, the sequence may come from powers, squares, triangular numbers, or step-by-step motion such as a frog making variable jumps. The key idea is to stop thinking about the full numbers and track only their remainders. ---

Learning Objectives

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Core Idea

This means that $a$ and $b$ have the same remainder upon division by $m$. Examples:

• $\qquad 29 \equiv 1 \pmod 7$

• $\qquad 38 \equiv 3 \pmod 5$

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Basic Rules of Congruence

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Sequence Patterns Modulo $m$

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Triangular Numbers and Frog Positions

So many frog problems reduce to studying $\qquad \dfrac{n(n+1)}{2} \pmod m$ ::: ---

Safe Periodicity Fact for Triangular Numbers

For example, modulo $7$, $\qquad T_n \equiv 0,1,3,6,3,1,0,\dots$ with period $7$. ---

Minimal Worked Examples

Example 1 Find the remainder when $\qquad 3^{17}$ is divided by $10$. The powers of $3$ modulo $10$ are: $\qquad 3,9,7,1,3,9,7,1,\dots$ This has period $4$. Since $\qquad 17 \equiv 1 \pmod 4$, we get $\qquad 3^{17} \equiv 3 \pmod{10}$ So the remainder is $\boxed{3}$. --- Example 2 Find all residues of $n$ modulo $7$ for which $\qquad \dfrac{n(n+1)}{2} \equiv 1 \pmod 7$ Compute $T_n$ modulo $7$ for $\qquad n=0,1,2,3,4,5,6$: $\qquad 0,1,3,6,3,1,0$ So the remainder $1$ occurs when $\qquad n \equiv 1 \text{ or } 5 \pmod 7$ --- Example 3 How many values of $n$ from $0$ to $20$ satisfy $\qquad \dfrac{n(n+1)}{2} \equiv 0 \pmod 3$? Since $\qquad 2$ is invertible modulo $3$, this is equivalent to $\qquad n(n+1)\equiv 0 \pmod 3$ So $\qquad n \equiv 0 \text{ or } 2 \pmod 3$ From $0$ to $20$, there are $21$ integers:

• $7$ with residue $0$

• $7$ with residue $2$

So the total is $\boxed{14}$. ---

Counting Terms in a Residue Class

This is exactly how many modular sequence counting questions are solved. ---

Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If $T_n=\dfrac{n(n+1)}{2}$, then $T_n \bmod 7$ for $n=0,1,2,3,4,5,6$ is" options=["$0,1,2,3,4,5,6$","$0,1,3,6,3,1,0$","$1,2,3,4,5,6,0$","$0,1,0,1,0,1,0$"] answer="B" hint="Compute directly for one full cycle." solution="Compute $\qquad T_n=\dfrac{n(n+1)}{2}$ for $n=0,1,2,3,4,5,6$. The remainders modulo $7$ are $\qquad 0,1,3,6,3,1,0$ So the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the remainder when $3^{17}$ is divided by $10$." answer="3" hint="Use the repeating last-digit pattern." solution="The powers of $3$ modulo $10$ repeat as $\qquad 3,9,7,1$ with period $4$. Since $\qquad 17 \equiv 1 \pmod 4$, we get $\qquad 3^{17}\equiv 3 \pmod{10}$ Hence the answer is $\boxed{3}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $a\equiv b \pmod m$, then $a+c\equiv b+c \pmod m$","If $a\equiv b \pmod m$, then $a^2\equiv b^2 \pmod m$","Division is always valid in congruences","A modular sequence often becomes periodic because only finitely many residues exist"] answer="A,B,D" hint="Use the standard rules of congruence carefully." solution="1. True.

True.

False. Division is not automatically valid modulo $m$.

True. There are only finitely many residue classes modulo $m$.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="How many integers $n$ with $0\le n\le 20$ satisfy $\dfrac{n(n+1)}{2}\equiv 0 \pmod 3$?" answer="$14$" hint="Use $n(n+1)\equiv 0 \pmod 3$." solution="We need $\qquad \dfrac{n(n+1)}{2}\equiv 0 \pmod 3$ Since $2$ is invertible modulo $3$, this is equivalent to $\qquad n(n+1)\equiv 0 \pmod 3$ So $\qquad n\equiv 0 \text{ or } 2 \pmod 3$ From $0$ to $20$, there are $21$ integers, with residues mod $3$ distributed equally:

• $7$ numbers congruent to $0$

• $7$ numbers congruent to $2$

Hence the total count is $\qquad 7+7=14$ Therefore the answer is $\boxed{14}$." ::: ---

Summary

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Part 4: Linear congruences

Linear Conguences

Overview

A linear congruence is an expression of the form $\qquad ax \equiv b \pmod m$ where the goal is to find all integers $x$ satisfying the congruence. In CMI-style questions, this topic is not just about plugging numbers into a formula. It tests when solutions exist, how many solutions exist, how to reduce a congruence, how to use inverses modulo $m$, and how to construct numbers satisfying several congruence conditions at once. ---

Learning Objectives

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Core Idea

To solve it, we usually reduce it to a simpler congruence or multiply by the inverse of $a$ modulo $m$ if that inverse exists. ---

Basic Meaning

Equivalently, $\qquad ax-b = km$ for some integer $k$. ---

Existence Criterion

This is the single most important solvability criterion in the topic. ---

Number of Solutions

If $d\nmid b$, then there are no solutions. ---

Coprime Case

So in the coprime case, solving a linear congruence is basically the same as dividing by $a$ modulo $m$. ---

Modular Inverse

Such an inverse exists if and only if $\qquad \gcd(a,m)=1$ ::: Example Since $\qquad 3\cdot 5 = 15 \equiv 1 \pmod 7$ the inverse of $3$ modulo $7$ is $5$. ---

Reducing a Congruence

This reduced congruence has coprime coefficient and modulus, so it can be solved by inversion. ---

Standard Solving Procedure

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Minimal Worked Examples

Example 1 Solve $\qquad 3x \equiv 2 \pmod 7$ Since $\gcd(3,7)=1$, inverse of $3$ modulo $7$ exists. Since $\qquad 3^{-1} \equiv 5 \pmod 7$ we get $\qquad x \equiv 5\cdot 2 = 10 \equiv 3 \pmod 7$ So the solution is $\qquad x \equiv 3 \pmod 7$ --- Example 2 Solve $\qquad 6x \equiv 8 \pmod{14}$ Now $\qquad \gcd(6,14)=2$ and $2\mid 8$, so solutions exist. Divide by $2$: $\qquad 3x \equiv 4 \pmod 7$ The inverse of $3$ modulo $7$ is $5$, so $\qquad x \equiv 5\cdot 4 = 20 \equiv 6 \pmod 7$ Thus modulo $14$, the solutions are $\qquad x \equiv 6,\ 13 \pmod{14}$ There are exactly $2$ solutions modulo $14$, matching the gcd. --- Example 3 Solve $\qquad 4x \equiv 3 \pmod 6$ Since $\qquad \gcd(4,6)=2$ but $2\nmid 3$, no solution exists. ---

Chinese Remainder Viewpoint

This is exactly the pattern of many construction-type PYQs. ---

Building Special Residues

Suppose $\gcd(m,n)=1$. We often want:

• $a \equiv 1 \pmod m$ and $a \equiv 0 \pmod n$

• $b \equiv 0 \pmod m$ and $b \equiv 1 \pmod n$

Then for any target remainders $r,s$, the number $\qquad x \equiv ra + sb \pmod{mn}$ satisfies $\qquad x \equiv r \pmod m,\qquad x \equiv s \pmod n$ ::: This is extremely efficient in constructive problems. ---

Euclidean Algorithm Link

This is the theoretical reason inverses exist in the coprime case. ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The congruence $5x \equiv 1 \pmod 7$ has solution" options=["$x \equiv 1 \pmod 7$","$x \equiv 2 \pmod 7$","$x \equiv 3 \pmod 7$","$x \equiv 5 \pmod 7$"] answer="C" hint="Find the inverse of $5$ modulo $7$." solution="Since $5\cdot 3=15\equiv 1\pmod 7$, the inverse of $5$ modulo $7$ is $3$. Therefore $\qquad x \equiv 3\cdot 1 \equiv 3 \pmod 7$ Hence the correct option is $\boxed{C}$." ::: :::question type="NAT" question="How many solutions modulo $12$ does the congruence $4x \equiv 8 \pmod{12}$ have?" answer="4" hint="Use $\gcd(a,m)$." solution="Here $\qquad \gcd(4,12)=4$ and $4\mid 8$, so the congruence has solutions. The number of solutions modulo $12$ is exactly $\gcd(4,12)=4$. Hence the answer is $\boxed{4}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["The congruence $ax\equiv b\pmod m$ has a solution iff $\gcd(a,m)\mid b$","If $\gcd(a,m)=1$, then $a$ has an inverse modulo $m$","The congruence $4x\equiv 3\pmod 6$ has a solution","If $\gcd(a,m)=d$ and $d\mid b$, then $ax\equiv b\pmod m$ has exactly $d$ solutions modulo $m$"] answer="A,B,D" hint="Use the standard theory for linear congruences." solution="1. True.

True.

False, because $\gcd(4,6)=2$ and $2\nmid 3$.

True.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Solve $6x \equiv 8 \pmod{14}$." answer="$x \equiv 6,13 \pmod{14}$" hint="First divide by the gcd." solution="We have $\qquad \gcd(6,14)=2$ and $2\mid 8$, so solutions exist. Divide by $2$: $\qquad 3x \equiv 4 \pmod 7$ The inverse of $3$ modulo $7$ is $5$, so $\qquad x \equiv 5\cdot 4 = 20 \equiv 6 \pmod 7$ Thus modulo $14$, the solutions are $\qquad x \equiv 6,\ 13 \pmod{14}$ Hence the complete solution set modulo $14$ is $\boxed{x \equiv 6,13 \pmod{14}}$." ::: ---

Summary

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="What is the remainder when $3^{2023}$ is divided by 7?" options=["1","2","3","4"] answer="3" hint="Identify the cycle length of powers of 3 modulo 7." solution="We compute the first few powers of 3 modulo 7:
$3^1 \equiv 3 \pmod 7$
$3^2 \equiv 9 \equiv 2 \pmod 7$
$3^3 \equiv 3 \cdot 2 \equiv 6 \pmod 7$
$3^4 \equiv 3 \cdot 6 \equiv 18 \equiv 4 \pmod 7$
$3^5 \equiv 3 \cdot 4 \equiv 12 \equiv 5 \pmod 7$
$3^6 \equiv 3 \cdot 5 \equiv 15 \equiv 1 \pmod 7$
The cycle length is 6. We divide the exponent 2023 by 6:
$2023 = 6 \times 337 + 1$
Therefore, $3^{2023} \equiv (3^6)^{337} \cdot 3^1 \equiv 1^{337} \cdot 3 \equiv 3 \pmod 7$.
The remainder is 3."
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:::question type="NAT" question="Find the smallest positive integer $x$ that satisfies $13x \equiv 7 \pmod{20}$." answer="19" hint="Find the multiplicative inverse of 13 modulo 20 using the Extended Euclidean Algorithm." solution="We need to solve $13x \equiv 7 \pmod{20}$.
First, find the inverse of $13 \pmod{20}$. Using the Extended Euclidean Algorithm:
$20 = 1 \cdot 13 + 7$
$13 = 1 \cdot 7 + 6$
$7 = 1 \cdot 6 + 1$
Working backwards to express 1 as a linear combination of 13 and 20:
$1 = 7 - 1 \cdot 6$
$1 = 7 - 1 \cdot (13 - 1 \cdot 7)$
$1 = 2 \cdot 7 - 1 \cdot 13$
$1 = 2 \cdot (20 - 1 \cdot 13) - 1 \cdot 13$
$1 = 2 \cdot 20 - 3 \cdot 13$
So, $-3 \cdot 13 \equiv 1 \pmod{20}$.
Since $-3 \equiv 17 \pmod{20}$, the inverse of $13 \pmod{20}$ is 17.
Now multiply both sides of the congruence by 17:
$17 \cdot 13x \equiv 17 \cdot 7 \pmod{20}$
$1x \equiv 119 \pmod{20}$
$119 = 5 \cdot 20 + 19$, so $119 \equiv 19 \pmod{20}$.
Thus, $x \equiv 19 \pmod{20}$.
The smallest positive integer $x$ is 19."
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:::question type="MCQ" question="Which of the following statements about the linear congruence $ax \equiv b \pmod n$ is FALSE?" options=["If $\operatorname{gcd}(a, n) = 1$, then there is a unique solution modulo $n$.","If $\operatorname{gcd}(a, n) = d$, then solutions exist if and only if $d \mid b$.","If $n$ is a prime number and $a \not\equiv 0 \pmod n$, then a solution always exists.","If $\operatorname{gcd}(a, n) > 1$, then there are no solutions."] answer="If $\operatorname{gcd}(a, n) > 1$, then there are no solutions." hint="Consider cases where $\operatorname{gcd}(a, n) > 1$ but solutions still exist." solution="Let's evaluate each option:
* If $\operatorname{gcd}(a, n) = 1$, then there is a unique solution modulo $n$. This is TRUE. If $\operatorname{gcd}(a, n) = 1$, $a$ has a unique inverse $a^{-1} \pmod n$, so $x \equiv ba^{-1} \pmod n$ is the unique solution.
* If $\operatorname{gcd}(a, n) = d$, then solutions exist if and only if $d \mid b$. This is TRUE. This is the fundamental existence condition for linear congruences.
* If $n$ is a prime number and $a \not\equiv 0 \pmod n$, then a solution always exists. This is TRUE. If $n$ is prime and $a \not\equiv 0 \pmod n$, then $\operatorname{gcd}(a, n) = 1$. By the first statement, a unique solution exists.
* If $\operatorname{gcd}(a, n) > 1$, then there are no solutions. This is FALSE. For example, consider $2x \equiv 2 \pmod 4$. Here, $\operatorname{gcd}(2, 4) = 2 > 1$. However, $2 \mid 2$, so solutions exist. Specifically, $x=1$ and $x=3$ are solutions modulo 4. The statement should be: "If $\operatorname{gcd}(a, n) > 1$ and $\operatorname{gcd}(a, n) \nmid b$, then there are no solutions."
Therefore, the false statement is "If $\operatorname{gcd}(a, n) > 1$, then there are no solutions.""
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:::question type="NAT" question="Find the largest two-digit integer $N$ such that $N \equiv 2 \pmod 5$ and $N \equiv 3 \pmod 7$." answer="87" hint="Formulate a system of congruences and solve for $N$ using substitution or the Chinese Remainder Theorem." solution="We are given the system of congruences:

$N \equiv 2 \pmod 5$

$N \equiv 3 \pmod 7$

From (1), $N$ can be written as $N = 5k + 2$ for some integer $k$.
Substitute this into (2):
$5k + 2 \equiv 3 \pmod 7$
$5k \equiv 1 \pmod 7$
To find $k$, we need the inverse of $5 \pmod 7$. We can see that $5 \cdot 3 = 15 \equiv 1 \pmod 7$. So, the inverse of $5 \pmod 7$ is 3.
Multiply both sides by 3:
$3 \cdot 5k \equiv 3 \cdot 1 \pmod 7$
$15k \equiv 3 \pmod 7$
$k \equiv 3 \pmod 7$
So, $k$ can be written as $k = 7j + 3$ for some integer $j$.
Substitute this expression for $k$ back into the equation for $N$:
$N = 5(7j + 3) + 2$
$N = 35j + 15 + 2$
$N = 35j + 17$
We are looking for the largest two-digit integer $N$.
If $j=0$, $N = 17$.
If $j=1$, $N = 35(1) + 17 = 52$.
If $j=2$, $N = 35(2) + 17 = 70 + 17 = 87$.
If $j=3$, $N = 35(3) + 17 = 105 + 17 = 122$, which is a three-digit number.
Thus, the largest two-digit integer $N$ satisfying the conditions is 87."
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What's Next?