Solution for this problem:
Step 1: Understanding the Problem
We need to find all positive integers \(n\) such that:
\(2^n−1≡0\) \((mod 7)\)
This means that \(2^n≡1\) \((mod 7)\).
Step 2:Finding the Order of 2 Modulo 7
The order of \(2\) modulo \(7\) is the smallest positive integer \(d\) such that:
\(2^d≡1\) \((mod7)\)
We calculate powers of \(2\) modulo \(7\):
- \(2^1≡2\) \((mod 7)\)
- \(2^2≡4\)\((mod 7)\)
- \(2^3≡8≡1\) \((mod 7)\)
Since \(2^3≡1\) \((mod 7)\), the order of \(2\) modulo \(7\) is \(3\). This means that \(2^n≡1\) \((mod 7)\) if and only if \(n\) is a multiple of \(3\).
Step 3: Finding Valid n
Since \(n\) must be a multiple of \(3\), the set of all positive integers that satisfy the condition is:
\(n=3k\), \(k∈Z+\)
Final Answer:
The positive integers \(n\) that satisfy the given condition are:
\(n=3,6,9,12,…\)
or, in general:
\(n=3k\) for \(k∈Z+\).
This means that \(3\) must be any positive multiple of \(3\).
Understanding Modulo 7 (mod 7)
The modulo operation (mod) finds the remainder when one number is divided by another. Specifically, mod 7 means dividing a number by \(7\) and keeping only the remainder.
How Mod 7 Works
For any integer \(a\), we write:\(a\) \(mod 7\)= \(r\)
where \(r\) is the remainder when \(a\) is divided by \(7\). The remainder \(r\) is always in the range:
\(0,1,2,3,4,5,or 6\)
Examples of Mod 7
- \(10\) \(mod 7\)= \(3\) (since \(10÷7=1\) remainder 3)
- \(15\) \(mod 7\)= \(1\) (since \(15÷7=2\) remainder 1)
- \(21\) \(mod 7\)= \(0\) (since \(21÷7=3\) remainder 0)
Application in the Problem
In our problem, we check when: \(2^n−1≡0\) \((mod7)\)
This means \(2^n\) must be congruent to \(1\) modulo \(7\). By calculating different powers of \(2\) modulo \(7\), we found:
- \(2^3≡1\) \((mod 7)\)
This tells us that \(n\) must be a multiple of \(3\).
Conclusion
Mod 7 helps in number theory by simplifying problems involving divisibility. It’s widely used in competitive math, cryptography, and computer science! 🚀
Also Read About : “Unique IMO 1962 Puzzle Solved: Finding Smallest Number Digit Rotation Trick”
Khairun
