Solution: We are asked to count how many of the first 100 positive integers satisfy the congruence: - Coaching Toolbox
Understanding and Solving: Counting How Many of the First 100 Positive Integers Satisfy a Given 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.
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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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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
