The number of elements in the set n N:10 n 100 and 3^n-3 is a multiple of 7 is ___________…

Question

The number of elements in the set

${n\in N:10\le n\le 100$ and ${3}^{n}-3$ is a multiple of 7$}$ is \_\_\_\_\_\_\_\_\_\_\_.

Answer: 15

Accepted Answer

To determine the number of elements in the given set, we need to find how many natural numbers $n$ between $10$ and $100$ (inclusive) satisfy the condition that ${3}^{n}-3$ is a multiple of $7$.

Recall that for any integers $a$ and $b$, $a$ is a multiple of $b$ if there exists an integer $k$ such that $a=bk$. So in our case, we need to find how many $n$ satisfy the equation ${3}^{n}-3=7k$ for some integer $k$.

Notice that ${3}^{n}-3=3({3}^{n-1}-1)$. We want this expression to be a multiple of 7. Let's explore a few powers of 3 modulo 7:

${3}^{1}\equiv 3\, (\bmod \, 7)$

${3}^{2}\equiv 9\equiv 2\, (\bmod \, 7)$

${3}^{3}\equiv 27\equiv 6\, (\bmod \, 7)$

${3}^{4}\equiv 81\equiv 4\, (\bmod \, 7)$

${3}^{5}\equiv 243\equiv 5\, (\bmod \, 7)$

${3}^{6}\equiv 729\equiv 1\, (\bmod \, 7)$

We observe that ${3}^{n}\, (\bmod \, 7)$ follows a cycle of length 6. So, ${3}^{n-1}\, (\bmod \, 7)$ also follows the same cycle, but shifted:

${3}^{0}\equiv 1\, (\bmod \, 7)$

${3}^{1}\equiv 3\, (\bmod \, 7)$

${3}^{2}\equiv 2\, (\bmod \, 7)$

${3}^{3}\equiv 6\, (\bmod \, 7)$

${3}^{4}\equiv 4\, (\bmod \, 7)$

${3}^{5}\equiv 5\, (\bmod \, 7)$

We want $3({3}^{n-1}-1)\equiv 0\, (\bmod \, 7)$, which means that ${3}^{n-1}-1\equiv 0\, (\bmod \, 7)$. From the cycle above, we see that this is true when $n-1$ is a multiple of 6, or equivalently, when $n$ is one more than a multiple of 6.

Now let's find the multiples of 6 between 10 and 100:

$12,18,24,30,36,42,48,54,60,66,72,78,84,90,96$

Adding 1 to each of these values, we get the set of natural numbers $n$ that satisfy the given condition:

$13,19,25,31,37,43,49,55,61,67,73,79,85,91,97$

There are 15 elements in this set. Therefore, the number of elements in the given set is $15$.

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