Long division

Some time back in the olden days, @robeastaway posted this:

Before I say anything else: I do not consider this a useful question for 11-year-olds or anyone else. I particularly dislike that a method is mandated - long division is a fine algorithm, but I don't think it's appropriate to say "you must do sums precisely this way."

That is not maths. That is computer programming, with students as the computers.

With that off my chest, I thought I might have a look at some of the possible methods for calculating 1,118 ÷ 43.

Long division

You can do it by long division, of course. It may even be your preference.

Here's how it works:

       ______2_6_
    43 ) 1 1 1 8
         0 | | |    No 43s go into 1.
         =======
         1 1 | |
           0 | |    No 43s go into 11
         =======
         1 1 1 |
           8 6 |    Finally! 2 × 43 = 86, so we can write 2 above the last 1.
         =======
           2 5 8     
           2 5 8    6 × 43 = 258, so we can put 6 above the 8.
           =====
               0    No remainder, and we're done.

That tells us the answer is 26 - but there's a lot that can go wrong there.

Estimate and check

The sum is "a bit more than 1000 divided by a bit more than 40", so an answer in the region of 25 seems plausible.

Working out 43 × 25 gives 1075, which is 43 short of the target. That means we must need 43 × 26.

Who threw that?!

"Forty-three sevens is 301," pointed out the Mathematical Ninja. "So you can work out 7826÷301, which is very clearly 26."

And the Mathematical Ninja was gone.

Algebra

(I know, this might not be perfect for 11-year-olds.)

Let the answer be $10a + b$, since it's clearly between 10 and 100.

Then $(40+3)(10a+b) = 1118$.

Expanding the brackets gives $400a + (30a + 40b) + 3b$.

We can surmise that $a=2$, because 3 is too high (we'd get 1200 and change), and 1 far too low. That makes the whole thing $800 + 60 + 40b + 3b = 1118$, or $40b + 3b = 258$.

Now, if $b$ were 6, it would make $40b = 240$, which is in the right ballpark - in fact, it's bang on.

(If we knew that the answer was an integer, we could state immediately that $b=6$ - only numbers that end in 6 give a final digit of 8 when multiplied by 3.)

Chunking

"$2 \times 43 = 86$, so $20 \times 43 = 860$. Take that off, we've still got 258. Now, $5\times 43 = 215$, take that off and we've still got 43 left over. That's $1\times 43$, so it's $26 \times 43$."

How a normal person would do it

How a normal person works it out

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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