Key Concepts and Formulas

• Modular Arithmetic: The concept of congruence, $a \equiv b \pmod{m}$, means that $a$ and $b$ have the same remainder when divided by $m$. This is equivalent to $a-b$ being a multiple of $m$.
• Cyclicity of Powers: For any integer $a$ and modulus $m$, the sequence of powers $a^1 \pmod{m}, a^2 \pmod{m}, a^3 \pmod{m}, \ldots$ is periodic.
• Counting Integers in a Range: The number of integers in a closed interval $[a, b]$ is $b - a + 1$.

Step-by-Step Solution

Step 1: Translate the problem into a modular congruence. The problem states that $3^n - 3$ is a multiple of $7$. This can be written in terms of modular arithmetic as: $3^n - 3 \equiv 0 \pmod{7}$ Adding $3$ to both sides, we get: $3^n \equiv 3 \pmod{7}$ We need to find the number of natural numbers $n$ in the range $10 \leq n \leq 100$ that satisfy this congruence.

Step 2: Determine the cycle of powers of $3$ modulo $7$. We calculate the first few powers of $3$ modulo $7$: $3^1 \equiv 3 \pmod{7}$ $3^2 \equiv 9 \equiv 2 \pmod{7}$ $3^3 \equiv 3^2 \cdot 3 \equiv 2 \cdot 3 \equiv 6 \pmod{7}$ $3^4 \equiv 3^3 \cdot 3 \equiv 6 \cdot 3 \equiv 18 \equiv 4 \pmod{7}$ $3^5 \equiv 3^4 \cdot 3 \equiv 4 \cdot 3 \equiv 12 \equiv 5 \pmod{7}$ $3^6 \equiv 3^5 \cdot 3 \equiv 5 \cdot 3 \equiv 15 \equiv 1 \pmod{7}$ $3^7 \equiv 3^6 \cdot 3 \equiv 1 \cdot 3 \equiv 3 \pmod{7}$ The sequence of remainders is $(3, 2, 6, 4, 5, 1)$, and it repeats every $6$ terms. The cycle length is $6$.

Step 3: Identify the condition on $n$ for the congruence to hold. We are looking for $3^n \equiv 3 \pmod{7}$. From our calculations, this occurs when $n=1, 7, 13, \ldots$. This pattern indicates that the exponent $n$ must have a remainder of $1$ when divided by the cycle length, which is $6$. Therefore, the condition on $n$ is: $n \equiv 1 \pmod{6}$ This means $n$ can be expressed in the form $n = 6k + 1$, where $k$ is an integer. Since $n$ is a natural number, $n \geq 1$, which implies $6k+1 \geq 1$, so $6k \geq 0$, meaning $k \geq 0$.

Step 4: Apply the given range for $n$. We are given that $10 \leq n \leq 100$. Substituting $n = 6k + 1$: $10 \leq 6k + 1 \leq 100$ Subtract $1$ from all parts of the inequality: $10 - 1 \leq 6k \leq 100 - 1$ $9 \leq 6k \leq 99$ Divide all parts by $6$: $\frac{9}{6} \leq k \leq \frac{99}{6}$ $1.5 \leq k \leq 16.5$ Since $k$ must be an integer, the possible values for $k$ are $2, 3, 4, \ldots, 16$.

Step 5: Count the number of valid values of $k$. The number of integer values for $k$ in the range $[2, 16]$ is given by: Number of values $= (\text{Last value}) - (\text{First value}) + 1$ Number of values $= 16 - 2 + 1 = 15$. Each valid integer value of $k$ corresponds to a unique value of $n$ that satisfies all the conditions of the problem.

Common Mistakes & Tips

• Incorrect Cycle Identification: Carefully compute the first few powers modulo $m$ to ensure the correct cycle length is identified. A mistake here will lead to an incorrect general form for $n$.
• Off-by-One Errors in Counting: When counting the number of integers in a range, remember to add $1$ to the difference between the last and first integer.
• Misinterpreting the Range for $k$: The initial condition $n \geq 1$ implies $k \geq 0$. However, the specific range for $n$ ($10 \leq n \leq 100$) might restrict $k$ to a higher starting value, as seen in Step 4.

Summary The problem requires finding the number of integers $n$ in the range $[10, 100]$ such that $3^n - 3$ is divisible by $7$. This translates to the congruence $3^n \equiv 3 \pmod{7}$. By examining the powers of $3$ modulo $7$, we found that the congruence holds when $n \equiv 1 \pmod{6}$. Expressing $n$ as $6k+1$ and applying the given range for $n$, we determined the possible integer values for $k$ to be from $2$ to $16$. Counting these values, we found there are $15$ such integers.

The final answer is $\boxed{15}$.