MORE ON MULTIPLES

Recall from Practice with Multiples that the multiples of a number $\,\,x\,\,$ are $\,\,x\,$, $\,\,2x\,$, $\,\,3x\,$, $\,\,4x\,$, and so on.
To test if something is a multiple of $\,\,x\,$, 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:

• Is $\,48\,$ a multiple of $\,6\,$?
Yes, since $\,6\,$ goes into $\,48\,$ evenly.
• Is $\,48\,$ a multiple of $\,7\,$?
No, since $\,7\,$ doesn't go into $\,48\,$ evenly.
• Is $\,2{,}652\,$ a multiple of $\,13\,$?
A calculator (or wolframalpha.com) is needed for this one:
$\,\frac{2652}{13} = 204\,$
So, $\,13\,$ goes into $\,2{,}652\,$ evenly ($\,204\,$ times), so $\,2{,}652\,$ is a multiple of $\,13\,$.
• Is $\,1,111,111,111,111,111,111,111\,$ a multiple of $\,3\,$?
Your calculator fails you here—the number is too big.
(But, wolframalpha.com can still handle it!)
However, no computing device is needed—the divisibility by $\,3\,$ test does the job effortlessly:
Since $\,\overset{\text{22 ones}}{\overbrace{1+1+\cdots+1}} = 22\,$, and since $\,22\,$ isn't divisible by $\,3\,$,
then $\,1,111,111,111,111,111,111,111\,$ isn't divisible by $\,3\,$.
(You might want to review these divisibility equivalences.)

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?
Absolutely!!

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 $\,55\,$.