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can some one explain with a clear example this theorem for me,

Let ($A_1$, $A_2$, $A_3$,..., $A_n$) be integars and $p$ a prime number.

if $p|(A_1A_2A_3...A_n)$ then there exist some $1 \leq k \leq N $ such that $p|A_k$.

Then there is a example on this which says,

is 6^100 divisible by 64? answer is yes, why is this? 64 is not a prime number the theorem says p must be a prime number?

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Counter example : $$3|111$$ – lab bhattacharjee Apr 11 '14 at 11:37
sorry what do you mean? – user142405 Apr 11 '14 at 11:40
$3$ divides $$1\cdot10^2+10^1\cdot1+10^0\cdot1$$ – lab bhattacharjee Apr 11 '14 at 11:41
@labbhattacharjee I think the statement is that if $p$ divides a product of integers then it divides one of the integers themselves, not that if $p$ divides a number it divides each of the digits. – Alex J Best Apr 11 '14 at 11:42
where did the plus sign come from I think you have not understood the question the (A1A2A3...An) is a product I believe Ak is a term in the product – user142405 Apr 11 '14 at 11:44

This is a rather simple statement that is written in a pretty unclear manner.

What they are saying is that if p is a prime number and $p\vert ab \implies p\vert a$ or $p\vert b$

This can easily be proven with the following proposition:

If a,b,c are integers with $c\vert ab$ and $gcd(a,c)=1 \implies c\vert b$

As people have said in the comments, the example does not really relate to the theorem at all.

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