Long division taken apart

When we divide, we are trying to find out how many times the divisor (e.g. 21) goes into the dividend (e.g. 1543).

We could just repeatedly subtract, keeping a tally of the number of times we have subtracted, stopping when further subtraction would give a negative result.  At least, that is how we could do divmod division.

Keeping such a tally would be tedious, however.  It is better to subtract large but convenient multiples of the divisor, than smaller multiples, keeping tally of these separately as we go.

In our usual decimal system of numeration, powers of ten are particularly convenient multipliers.

In 1543, the highest nonzero column is the thousands column.  We can therefore try subtracting thousands of `21`s.  We could not subtract `21` thousand even once from `1` thousand and anything without the result being negative.  So we need `0` thousands of `21`s, and we still have `1` thousand and something.

Next, we try hundreds of `21`s.  Again, we cannot subtract even `21` hundreds even once from `15` hundred and something, without the result being negative.  So we need `0` hundreds of `21`s, and we still have `15` hundred and something.

Next, we try tens of `21`s.  We can subtract `21` tens from `154` tens and something.  Indeed, we can subtract `7` times, a total of `147` tens, without getting negative result.  So now we have that `154` tens and `3` is the same as `7` lots of `21` tens, and then `7` tens and `3`.

Next we try whole `21`s.  We can subtract `21` from `73` just `3` times without going negative.  This subtracts a total of `63` from `73`, leaving `10`.

So now we have that `1543` is the same as `7` lots of `21` tens, and `3` lots of `21` ones, and a further `10`.

This can instead be understood as `70` lots of `21`, and `3` lots of `21`, and `10` more.  This in turn is `73 xx 21` and `10`.  i.e.

`\qquad\qquad\qquad 1543 = (73 xx 21) + 10.`

If we are doing divmod division, we can stop :

`\qquad\qquad\qquad 1543 -: 21 = 73" remainder "10.`

But if, instead of a remainder, we want the quotient to have a fractional part expressed as a decimal, then we keep going.

We have `10` that still needs to be divided by `21`.  So now we try subtracting tenths of `21`.  `10` is `100` tenths.  We can subtract `4` lots of tenths of `21`, i.e. `84` tenths, from `100` tenths, leaving `16` tenths.

Next, we try subtracting hundredths of `21` from `160` hundredths.  `7` hundredths of `21` is `147` hundredths, which when subtracted from `160` hundredths leaves `13` hundredths.

Stopping at this point, we find that

`\qquad\qquad\qquad 1543 = (73.47 xx 21) + 0.13`,

which we can write

`\qquad\qquad\qquad 1543 -: 21 = 73.47" remainder "0.13`.

The preceding discussion has been rather lengthy--such work is usually set out in a much more compressed format.  Unfortunately, long division seems to have been taught procedurally without adequate preparation, so that the elided computational form entrains elided thinking.  Even among the few in these calculator-infested days who can still actually do long division, a substantial fraction can give no convincing account of why the standard long division algorithm works.