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

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.

Understanding the Context

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.

Solved: 10) How many positive integers x

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Key Insights

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

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Final Thoughts

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