Products, sums and divisors

To prove that the pattern does not work for any $n$ , it is enough to give one counterexample.

$6\times6\times6\times6\times6$ is not divisible by $6+6+6+6+6=5\times6$ because it is not a multiple of $5$ .

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The reason it seemed to work was because $6+6=2\times6$ and $2$ is a divisor of $6$ . As is $3$ in $6+6+6$ .

Next $6+6+6+6=4\times6$ , and $4$ is a divisor of $6\times6$ . It is $5$ where it finally does not work.

The statement is:

Is it true that $6^n$ is divisible by $6n$ for a positive integer $n$ ?

Let $A = \frac{6^n}{6n}$ . We're going to prove that $A$ is not always an integer. $A = \frac{6^n}{6n}$ $\Rightarrow A = \frac{ 6^{n-1} }{n}$
$\Rightarrow A = \frac{ 6^{n-1} }{n} = \frac{ 2^{n-1}\cdot 3^{n-1} }{n}$

Now we see that; for $A$ to always be an integer, $n = 2^a, 3^b$ or $2^a \times 3^b$ for some $a,b \le n-1$ because if so, it will divide with the $2^{n-1}$ or $3^{n-1}$ or $6^{n-1}$ on the numerator. Since $n$ is of all positive integers, it can not just be of the form $n = 2^a, 3^b$ or $2^a \times 3^b$ . An example would be $n = 5$

Therefore, $6^n$ is not divisible by $6n$ for $\forall n \in \mathbb{Z}^+$

In general, the pattern will not work for the prime number of times excluding 2 and 3.

6^(prime number of times other then 2 and 3) will not divisible by (prime number of times other then 2 and 3)*6

we don't have to go far,

$6^1 ,6^2 ,6^3 ,6^4$ all are divisible according to the condition.

but, $\frac{6^5}{6+6+6+6+6}$ = $\frac{1296}{5}$

as, $6^5$ is not multiple of 5,so it is not divisible by 30(multiple of 5)

Counterexample: $n = 5$ . Hence, $\boxed{\text{No}}$ .

The prime factors of $6^n$ are only 2 and 3 for any n; For any number n which has a different prime factor (e.g. n = 5, 7, 10, 11, 14,...) the number $n \times 6$ has this prime factor as well, so $6^n$ is not divisible by it in those cases.

6^n-1/(n) is not for every positive integer

Not even close. The divisor is 6 x n Powers of 6 have only factors 2 and 3. The power of 6 will stop being divisible by 6 x n as soon as n has a factor that is not 2 or 3, at n=5.

Up to 4 terms it seems favorable but as we calculate with 5 terms, we the answer NO

6 x 6 x 6 x 6 x 6 and 6 + 6 + 6 + 6 + 6

6 x 6 x 6 x 6 x 6 and 6 ( 1 + 1 + 1 + 1 + 1 )

6 x 6 x 6 x 6 x 6 and 6 x 5

Hence, answer is NO

This is actually a poorly worded question. As the question is phrased the answer is yes. If the words 'is divisible' in the question were reworded to 'is evenly divisible' then the answer would be no.

Any number is not divisible by a prime number not in its prime factorisation. Here, for a number not,we are basically checking 6^n/(6n). We get 6^(n-1)/n. =2^(n-1)*3^(n-1)/n. If n is a prime number other than 2 or 3, it is indivisible.

Here we have the form 6×6×6.....(n times)/(6+6+6......{n times}) so, we can simplify it to 6^n/(6n) so here we can see it is written as 6^(n-1)/n so if 6^(n-1) is divisible by n then it is true but 6^(n-1) is not always divisible by n hence the answer is NO