Proving that a number has at least 3 distinct prime factors.

Question

Let abc be a 3-digit natural number (written in base 10). Prove that the 6-digit number abcabc has at least three distinct prime factors.

I know that to prove that the 6-digit number has at least three distinct prime factors, it suffices to show that abc has three distinct prime factors. However, the example that my teacher gave me was this,

Since abcabc = 1001·abc, it suffices to show that 1001 has three distinct prime factors. In fact, 1001 = 7·11·13

I seem to misunderstand the concept of a 3-digit number written in base 10. As far as I know, the base 10 system is the 0-9 numbering system. If so, then how can a 3-digit natural number be 1001? What have I misunderstood here?

Answer 1

It is a simple calculuation. $1001 \times abc = 1000 \times abc + abc = abcabc$.

Answer 2

$1001\cdot abc = abcabc$ is a calculation. In other words 1001 and $abc$ are factors of every number of the form $abcabc$.

It is easy to see that: $1000\cdot abc = abc000$. And $abc000 + abc = abcabc$.

Answer 3

The notation on this one is confusing. $abc$ is really a shorthand for $10^2a + 10b + c$, with each of $a, b, c$ in the range from $0$ to $9$, and similarly $abcabc$ means $10^5a + 10^4b + 10^3c + 10^2a + 10b + c$. This number can also be written as $1001 \times (10^2a + 10b + c)$.

And since $1001 = 7 \times 11 \times 13$, this guarantees $10^5a + 10^4b + 10^3c + 10^2a + 10b + c$ has at least three distinct prime factors even if $10^2a + 10b + c$ contributes no "new" prime factors, e.g., $539$ and $539539$.