Prime Numbers

For all positive integral $n$ , the units digit of $6^n$ will be 6. Subtracting $1$ from this will cause the number to be divisible by $5$ . The only prime multiple of $5$ is $5$ itself, which is obtained when $n=1$ . Thus, there is only $\boxed{1}$ solution.

We can calculate this by finding the value and pattern of $6^{n}-1$ where we will take n as: $n=1, 2$ and $3$

$6^{1}-1 = 6-1 = 5,$ where $n=1$

$6^{2}-1 = 36-1 = 35,$ where $n=2$

$6^{3}-1 = 216-1 = 215,$ where $n=3$

We notice that apart from $5$ (which has 2 factors 1 and 5)every next result has more than 2 factors (1, 5 and the number itself, etc).

A prime number has only $2$ factors: $1$ and number itself.

Thus, we have only $\boxed{1}$ solution to this question, that is the prime number $\boxed{5}$

Thus, the answer is $\boxed{1}$

All the powers of six with exponent two or higher ends with number 6.As result,the number ends with 5,obviously a composite number.Only the value of n=1 satisfy.

For all positive integer values of n, 6^n ends in the digit 6. (This is the trick where you just multiply the last digit to get the last digit of the next in the geometric sequence).

This means that if you subtract 1, you get a positive integer which ends in the digit 5: a multiple of 5.

There for the only prime is when n=1, so one solution.

The $6^n-1$ can be written as $6^n-1^n$ . We know that:

$\displaystyle a^n-b^n=\sum_{k=0}^{n-1}a^k \cdot b^{n-1-k}$

$\displaystyle X(n) = 6^n-1^n=(6-1)\sum_{k=0}^{n-1}6^k \cdot 1^{n-1-k} =$ $\displaystyle 5 \sum_{k=0}^{n-1}6^k$

The $X$ will always be multiple of 5 which is a prime number, so $X$ will always be divisible by 5. So the last prime number of $X(n)$ is $5$ and $X(1)=5$ so $\boxed{n=1}$

Any power of 6 have digit at unit place 6 . 6^n -1 divisible by 5 . Hence 6^1-1 is answer. There exist only solution 1

Cyclicity of number 6 is 6 only . So for each n 6^n -1 always be multiple of 5. The only prime number we can get is 5