MORE ON MULTIPLES

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:

- 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) |

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 $\,55\,$.