Animated Solution for Mathematics - Sequence and Series: The number of elements in the set $\{n\in \mathbb{N}:10\le n\le 100\text{ and }{3}^n-3\text{ is a multiple of }7\}$ is

A Journey into Modular Arithmetic

Have you ever felt that some math problems are like secret codes? You look at the expression $3^n-3$ being a multiple of $7$, and it seems like a wall of symbols.

But beneath that surface lies a beautiful, rhythmic structure. Today, we are going to crack this code, not by brute force, but by listening to the heartbeat of the numbers themselves.

The problem asks us to find how many natural numbers $n$ between $10$ and $100$ satisfy the condition that $3^n-3$ is a multiple of $7$. In the language of mathematics, being a "multiple of $7$" is just another way of saying the remainder is zero when divided by $7$.

So, we write:

Rearranging this, we get the core of our investigation:

This is our target. We aren't looking for the value of $3^n$; we are looking for the values of $n$ that make $3^n$ leave a remainder of $3$ when divided by $7$.

Let's observe the powers of $3$ modulo $7$. This is where the magic happens. We calculate the remainders step by step:

For $n=1$: $3^1=3\equiv 3(\mathrm{mod}7)$. (Bingo! We found our first match.)

For $n=2$: $3^2=9\equiv 2(\mathrm{mod}7)$.

For $n=3$: $3^3=27\equiv 6(\mathrm{mod}7)$.

For $n=4$: $3^4=81\equiv 4(\mathrm{mod}7)$.

For $n=5$: $3^5=243\equiv 5(\mathrm{mod}7)$.

For $n=6$: $3^6=729\equiv 1(\mathrm{mod}7)$.

Look at that sequence of remainders: $3,2,6,4,5,1$. Once we hit $1$, the cycle resets.

Why? Because $3^7=3^6\times 3\equiv 1\times 3=3(\mathrm{mod}7)$. The pattern $3,2,6,4,5,1$ will repeat indefinitely every $6$ steps.

This is the "rhythm" I mentioned earlier. It is the fundamental geometric reality of powers in modular arithmetic.

We need $3^n\equiv 3(\mathrm{mod}7)$. Looking at our cycle, this only happens when $n$ is in the first position of the cycle.

This means $n$ must be $1,7,13,19,\dots$. We can capture this entire sequence with a single, elegant algebraic expression:

where $k$ is any non-negative integer. This formula is our key. It maps every possible value of $n$ that satisfies our condition.

Now, we must respect the boundaries set by the problem: $10\le n\le 100$. We substitute our general form into this inequality:

Subtracting $1$ from all sides gives us:

Dividing by $6$ yields:

Simplifying these fractions, we get:

Since $k$ must be an integer (because it represents the number of cycles), the possible values for $k$ are the integers from $2$ to $16$ inclusive.

To find the number of elements, we simply count the integers in the set $\{2,3,\dots ,16\}$. The number of elements is given by the upper limit minus the lower limit, plus one:

There are exactly $15$ such numbers. We didn't need to calculate massive powers or perform complex divisions.

We simply identified the cycle, generalized the pattern, and constrained it within the given range. That is the elegance of number theory—it turns a mountain of calculation into a simple, beautiful dance of logic.