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Recall from Practice with Multiples that the multiples of a number [beautiful math coming... please be patient] $\,\,x\,\,$ are $\,\,x\,$, $\,\,2x\,$, $\,\,3x\,$, $\,\,4x\,$, and so on.
To test if something is a multiple of [beautiful math coming... please be patient] $\,\,x\,$, just see if $\,\,x\,\,$ goes into it evenly.

This is all fine and good when numbers are fairly small, or when we can call on a divisibility test for help.
For example, these questions are all pretty easy:

But now, try a question like this:
Is   $\,2^{57}3^{111}10^{42}\,$   divisible by $\,55\,$?
WolframAlpha can still help, if you interpret the result correctly.
But, can it be done without this kind of ‘super-power’ help?

The primes numbers come to the rescue!
Every number has a unique decomposition into prime factors.
(Read about primes and the unique prime decomposition on pages 108–109 of the text.)
To be a multiple of $\,55 = 5\cdot 11\,$, a number must have a prime factor of $\,5\,$ and a prime factor of $\,11\,$.
So, does the number   $\,2^{57}3^{111}10^{42}\,$   have these two prime factors?

We have to be careful—the number   $\,2^{57}3^{111}10^{42}\,$   isn't yet completely factored into primes.
So, let's do some renaming:

$\,2^{57}3^{111}10^{42}$($\,2\,$ and $\,3\,$ are prime, but $\,10\,$ isn't)
      $= \,2^{57}3^{111}(2\cdot 5)^{42}$(break $\,10\,$ into its prime factors)
      $= \,2^{57}3^{111}2^{42}5^{42}$(use an exponent law)
      $= \,2^{57+42}3^{111}5^{42}$(use an exponent law)
This is a much better name for that big number.
Now, we can see that it only has prime factors of $\,2\,$, $\,3\,$ and $\,5\,$.

So, is it a multiple of $\,55 = 5\cdot 11\,$?
It has the needed factor of $\,5\,$. (Indeed, it has forty-two factors of $\,5\,$.)
But, it doesn't have a factor of $\,11\,$.
So, it isn't a factor of $\,55\,$.