How many of the first 100 positive integers are congruent to 3 (mod 7)? - Moon Smoking
Title: How Many of the First 100 Positive Integers Are Congruent to 3 (mod 7)?
Title: How Many of the First 100 Positive Integers Are Congruent to 3 (mod 7)?
Understanding number modular arithmetic can reveal fascinating patterns—one interesting question is: How many of the first 100 positive integers are congruent to 3 modulo 7? This seemingly simple inquiry uncovers how evenly integers spread across residue classes and highlights the predictable patterns in modular systems.
Understanding the Context
What Does “Congruent to 3 mod 7” Mean?
When we say an integer n is congruent to 3 modulo 7, we write:
n ≡ 3 (mod 7)
This means n leaves a remainder of 3 when divided by 7. In other words, n can be expressed in the form:
n = 7k + 3
where k is a non-negative integer.
Image Gallery
Key Insights
Finding All Numbers ≤ 100 Such That n ≡ 3 (mod 7)
We want to count how many values of n in the range 1 to 100 satisfy n = 7k + 3.
Start by solving:
7k + 3 ≤ 100
7k ≤ 97
k ≤ 13.857…
Since k must be an integer, the largest possible value is k = 13.
Now generate the sequence:
For k = 0 → n = 7(0) + 3 = 3
k = 1 → n = 10
k = 2 → n = 17
...
k = 13 → n = 7(13) + 3 = 94 + 3 = 94 + 3? Wait: 7×13 = 91 → 91 + 3 = 94
🔗 Related Articles You Might Like:
📰 The sum of the known measurements is 2.1 + 2.7 + 3.0 + 2.4 = 10.2 meters. 📰 A museum curator is digitizing a collection of 120 historical telescopes, assigning each a unique digital ID starting from 001 to 120. If the average ID number of the first 90 telescopes is 45.3, what is the average ID number of the remaining 30 telescopes? 📰 A biostatistician analyzes health survey data from 400 low-income individuals and finds that 65% have limited access to fresh produce. Of those with limited access, 40% report chronic respiratory issues, while only 15% of those without limited access report such issues. What percentage of the entire group reports chronic respiratory issues?Final Thoughts
Wait: 7×13 = 91 → n = 91 + 3 = 94
k = 14 → 7×14 + 3 = 98 + 3 = 101 > 100 → too big
So valid values of k go from 0 to 13 inclusive → total of 14 values.
List the Numbers (Optional Verification):
The numbers are:
3, 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, 80, 87, 94
Count them — indeed 14 numbers.
Why Does This Pattern Occur?
In modular arithmetic with modulus 7, the possible residues are 0 through 6. When dividing 100 numbers, each residue class mod 7 appears approximately 100 ÷ 7 ≈ 14.28 times.
Specifically, residues 0 to 3 mod 7 occur 15 times in 1–98 (since 98 = 14×7), and then residues 4–6 only appear 14 times by 98. However, residue 3 continues into 100:
Indeed, n = 3, 10, ..., 94, and the next would be 101 — outside the range.
