Last week, in Fundamental Understanding of Mathematics XXXVIII, we took a look at a geometric interpretation of multiplication and division, using a number line. So far, all the examples have been small numbers, numbers that can be found in a standard multiplication table. Single digit multiplication and division that resolves into a single digit by single digit multiplication with a single digit remainder. With such simple problems, there's no need for a complicated division algorithm. As with addition and multiplication, though, as the numbers get larger, it makes sense to devise a method to allow us to use simple division fact families to solve complicated division problems.
Back in Fundamental Understanding of Mathematics XXVII I discussed an algorithm for dividing large numbers that relied on partial products. This is generally not taught in the American school system, and it looks strange, weird and scary to people brought up learning the American long division algorithm. However, since it does not rely on knowledge of the multiplication table, it might be a method that can be used by people who can add and subtract, but not multiply.
But I digress. This week we will take a look at multiple digit division, using the standard long division algorithm.
Just as the addition algorithm used repeated simple additions, and the multiplication algorithm used repeated simple multiplications, so the long division algorithm will use repeated simple divisions, with remainders.
The key to understanding long division is that we are not going to get "the answer" (since we will probably be dealing with a division problem that doesn't have a whole number answer) but we are going to get close to "the answer" and then add in a remainder. Let's look at an example: 49 divided by 5 has a quotient of 9 and a remainder of 4. We can see that our "answer," 9, isn't exact (9 x 5 isn't 49) but is close (it's only off by 4, the remainder.)
We write this as a math sentence:
How do we figure this out? We begin with what we know
The quotient, 9, is the largest possible multiple of 5 which is still less than 49. If we look at the multiples of five, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, we see that 45 is less than 49, but 50 is too large. 45 is the ninth multiple of 5, so
Notice that the remainder will be some number less than 5, since, if the remainder were greater than 5, we could use a higher multiple of 5. We find the remainder by doing the multiplication, then subtracting
If we take a look at a number line, we can see there are nine segments, five units long, and a shorter segment (the remainder) that's only four units long. We can see that the remainder segment will always be shorter than the other segments.
Let's look at a more complicated problem, 257 divided by 6. We'll stick with a single digit divisor, for now, but consider a problem with a two digit quotient. We will figure out each digit of that quotient separately, using simpler divisions.
First of all, how do we know that this problem has a two digit quotient? Consider the smallest possible 1 digit, 2 digit, 3 digit, 4 digit quotients: 1, 10, 100, 1000. If we try these out, we get the following:
10 times 6 is smaller than 257, and 100 times 6 is larger. So, the quotient will be greater than 10, a two digit number, and smaller than 100, a three digit number. Since the next smaller number than 100 is 99, a two digit number, the quotient will be a two digit number, somewhere between 10 and 99.
(This seems simplistic, but it's a good habit to get into. When we do division problems with multiple digit divisors, it's not always obvious how many digits the quotient will have. Suppose we are dividing by 327. Quickly multiplying 327 by powers of ten: 327, 3270, 32700, 327000, etc will show us how many digits the quotient will have. More on this when we discuss multiple digit divisors.)
So the quotient is somewhere between 10 and 99. Let's narrow that further by looking at how many 10s will be in the quotient.
50 is too big, so the quotient will be between 40 and 49. The tens digit will be a four.
We can now write the quotient using expanded notation, since we don't know the ones digit, we will replace that by an unknown: q. So quotient = 40 + q, where q is some one digit number.
Apply the distributive property
We can use the associative property to gather the unknowns
Subtract (40 x 6) from both sides
Let's organize this using the long division bracket
Notice that I'm subtracting 40 x 6, rather than 4 x 6, then "bringing down." Bringing down is a shortcut that obscures the fact that the "4" digit above the line represents 40.
We still don't know what the ones digit is. But, we have another simple division problem.
Earlier, we specified that q was a single digit number, so we know it is between 1 and 9. If we look at multiples of 6...
we see that 3 x 6 is too large, so q must be 2.
Using the long division bracket, we write the 2 in the one's place, multiply and subtract to find the remainder.
So, we can again devise an algorithm using our "divide and conquer" strategy that worked for addition and multiplication. In effect, we have discovered that
where 40 gives us the tens digit of our quotient, and 2 is the singles digit.
Have fun in the comments.