Solution: We are asked to count how many of the first 100 positive integers satisfy the congruence:

February 22, 2026

Understanding and Solving: Counting How Many of the First 100 Positive Integers Satisfy a Given Congruence

When it comes to number theory in mathematics, congruences play a vital role—especially in problems involving modular arithmetic. A common challenge often presented is: How many of the first 100 positive integers satisfy a particular congruence condition?

While the exact congruence isn’t specified, this article explores a general solution approach using modular arithmetic, walk through practical examples, and provides methods to efficiently count solutions within a finite range—such as the first 100 positive integers.

What Is a Congruence?

A congruence expresses whether two integers leave the same remainder when divided by a positive integer (the modulus). For example: x ≡ a (mod n) means that x and a leave the same remainder upon division by n, or equivalently, n divides (x − a).

In this context, we are interested in counting integers x in the set {1, 2, 3, ..., 100} such that: x ≡ a (mod n) for fixed integers a and n.

Example Problem

Let’s suppose the problem asks: How many of the first 100 positive integers are congruent to 3 modulo 7? That is, find the count of integers x such that: x ≡ 3 (mod 7), and 1 ≤ x ≤ 100

Step-by-Step Solution

Understand the Pattern of Solutions The general solution to x ≡ 3 (mod 7) is: x = 7k + 3, where k is any integer
Find Valid Values of k We need 1 ≤ 7k + 3 ≤ 100

Solve for k: 1 ≤ 7k + 3 ⇒ 7k ≥ –2 ⇒ k ≥ 0 (since k must be integer) 7k + 3 ≤ 100 ⇒ 7k ≤ 97 ⇒ k ≤ ⌊97/7⌋ = 13

So k ranges from 0 to 13 inclusive.

Count the Valid k Values k = 0, 1, 2, ..., 13 → total of 14 values

Thus, there are 14 integers between 1 and 100 that satisfy x ≡ 3 (mod 7).

General Strategy for Counting Solutions (1 ≤ x ≤ 100)

For a congruence x ≡ a (mod n), follow these steps:

Express solution set: x = n·k + a, where k is an integer
Apply bounds: 1 ≤ n·k + a ≤ 100 ⇒ (1 – a)/n ≤ k ≤ (100 – a)/n
Compute integer k values in this range: k_min = ⌈(1 – a)/n⌉ (round up) k_max = ⌊(100 – a)/n⌋ (round down) Number of valid k = k_max – k_min + 1 (if ≥ 0)

> Tip: Use ceiling and floor functions to avoid rounding errors.

If a > 100 or a < 1, no solutions exist in the range.

Another Common Example: Squares Modulo 4

Suppose instead the congruence condition is: How many of the squares of the first 100 positive integers satisfy: n² ≡ 1 (mod 4)?

Note: Squares mod 4 cycle in a predictable way:

If n is odd: n ≡ 1 or 3 mod 4 → n² ≡ 1 mod 4
If n is even: n ≡ 0 or 2 mod 4 → n² ≡ 0 mod 4

So n² ≡ 1 mod 4 iff n is odd.

Count odd integers from 1 to 100: 1, 3, 5, ..., 99 → total = 50.

Thus, 50 of the first 100 positive integers have squares congruent to 1 mod 4.

Why This Matters (Practical Applications)

Understanding how many numbers in a range satisfy a congruence is useful in:

Cryptography and coding theory
Algorithm design (e.g., hash functions, modular quadratic testing)
Exploring number patterns and puzzles

Summary

To solve “How many of the first 100 positive integers satisfy a congruence x ≡ a mod n?”:

Express x = n·k + a
Solve bounds for k: 1 ≤ n·k + a ≤ 100
Use algebra to find k_min and k_max
Count valid integers: k_max – k_min + 1 (if valid)

This method applies universally, whether a = 3, n = 7, or more complex quadratic residues.

Key Takeaways:

Congruences describe number patterns within modular frameworks
The first 100 integers offer a clean range for systematic testing
Algorithmic thinking simplifies counting solutions efficiently

For any specific congruence, apply modular arithmetic and integer bounds to count solutions precisely.

Try it yourself: Pick a modulus and remainder—count how many values from 1 to 100 satisfy x ≡ a (mod n) using the method above!