Some time ago I was making up a silly song about zero to the tune of Cherish by the Association. The original song goes like this:

Cherish is a word I use to describe
All the feeling that I have hiding here for you inside
You don't know how many times I've wished that I had told you
You don't know how many times I've wished that I could hold you
You don't know how many times I've wished that I could
Mold you into someone who could
Cherish me as much as I cherish you

My song, besides being silly, is about zero, so of course it started like this:

Zero is a word I use to designate
The nothingness that's left when I take two fours from eight.
I don't know how many times...

And then, because the creative juices were flowing, and it's a math song, what came out was:

I don't know how many times sixteen five-hundred twelve is.

So of course I had to have more lines like that. In order to scan properly, it had to have a two-syllable divisor followed by a four-syllable multiple of that divisor, and the lines had to rhyme. Only -teens would work because, while seven is two syllables, it sounds funny in the staccato rhythm of those lines. Thirteen, fourteen, fifteen, sixteen, eighteen, and nineteen are just fine. Seventeen is three syllables, but if I elide the first two together the way poets do making “even” into “e'en,” then I could also use it in a pinch.

The four-syllable multiple would have to end in twelve. No other numbers would rhyme. They would all have to be some-hundred twelve. This is a small enough number set that I could just try each one on a calculator, but I wanted a more elegant, more fun solution, so I decided to use backwards long division. To see how it works, let's start with some-hundred twelve divided by thirteen. I will use dots for the digits we don't know.

••
13 •12
  •• 
   •2
   •2

It should be apparent that the quotient has to end in four. It's the only number we can multiply by thirteen to get a number ending in two. So the last partial result must be fifty-two (4 × 13 = 52).

•4
13 •12
  •• 
   52
   52

The next partial result must end in six because it is the only digit we can subtract from one (eleven after borrowing) to get five. So the first digit of the quotient must be two, and the partial result must be twenty-six.

24
13 •12
  26 
   52
   52

From this it immediately follows that the first digit of the multiple is three.

24
13 312
  26 
   52
   52

I now have another potential line for my zero song.

I don't know how many times thirteen three-hundred twelve is.

Of course, I actually do know. That's what makes the song ironic.

Moving on to fourteen, we quickly run into trouble. The last partial result must be forty-two, but that leads to a first partial result that ends in seven. Since fourteen is even, all its multiples are also even. There is no partial result that ends in seven.

•3
14 •12
  •7 
   42
   42

But wait! The quotient doesn't have to end in three. It could also end in eight. However, now the last partial result has three digits, so we have to make some adjustments. In fact, we get an immediate result because 8 × 14 = 112, which also yields another potential line. But there is yet another.

•8
14 •12
  •• 
   112
   112

Plainly the first partial result must end in zero, which means that it is seventy.

58
14 812
  70 
   112
   112

This yields two more candidate lines.

I don't know how many times fourteen one-hundred twelve is.
I don't know how many times fourteen eight-hundred twelve is.

Someone may notice that 812 and 112 differ by 700, and 700 is the smallest multiple of fourteen that ends in 00. In fact, for each of the -teens, if we keep adding the smallest multiple that is also a multiple of 100, we will keep getting more numbers ending in twelve. In the case of thirteen, the smallest such multiple is 1300, so the number sequence begins at 312 and continues 1612, 2912, 4212, etc. Unfortunately, none of these numbers meet the four-syllable requirement for the song.

Fifteen yields no solution because every multiple of fifteen ends in zero or five. None can end in twelve.

Sixteen is the number I started with. It popped into my head because both sixteen and five-hundred twelve are powers of two (24 = 16 and 29 = 512). I knew immediately that five-hundred twelve was a multiple of sixteen. It is not the only one. Using the same technique I used for fourteen yields two more solutions: one-hundred twelve and nine-hundred twelve. This is because the least common multiple of 16 and 100 is 400, so numbers divisible by 16 and ending in 12 are 112, 512, 912, 1312, 1712, etc.

Here is the full list of candidate lines. I included seventeen, e'en though we are not in a pinch and don't need it.

I don't know how many times thirteen three-hundred twelve is.
I don't know how many times fourteen one-hundred twelve is.
I don't know how many times fourteen eight-hundred twelve is.
I don't know how many times sixteen one-hundred twelve is.
I don't know how many times sixteen five-hundred twelve is.
I don't know how many times sixteen nine-hundred twelve is.
I don't know how many times se'enteen six-hundred twelve is.
I don't know how many times eighteen six-hundred twelve is.
I don't know how many times nineteen nine-hundred twelve is.

I confess I was surprised to find so many, and more surprised to find the repetitions. Three numbers are multiples of -teens in two ways (112, 612, and 912).