For instance in the following problem:


To determine an initial 9 for the first number in the Quotient, you have to look at how many times does 48 go into 436 because; 48 doesn't go into 4 and it doesn't go into 43.

Then one might start trying to multiply 48 by increments of 1 or increments of 5 by hand until they find two numbers multiplied together to determine which number being multiplied gets one closest to the number without going over it; (this is awfully tedious and time consuming), for example: - 48*1 = 48, 48*2=96, 48*5=240 ... 48*9=432

The inefficiency of this method (by incrementing by one) can be displayed in a graph and spreadsheet (though when required to do these calculations by hand no such oversight is available):

enter image description here

Now one does not, in this example get hung up somewhere in the middle, since 9 as the first digit in the quotient is the end of the line, but if it was not, and the correct first digit of the quotient was something like 6, one would have to at the very least do the calculation for the first digit of the quotient 7 and also do the calculation to determine the first digit of 6, to determine that 6 is the correct number to use to reach the nearest result, since 7 multiplied by the divisor would be greater than the result.



What tools / methods can one use to estimate which two digits you need to try this with...without guessing / taking a shot in the dark, or calculating each and every possibility linearly?

  • 3
    $\begingroup$ Just a comment (i.e., not an answer!): Initially underestimating is a fine way to find the correct quotient, and is, in particular, the technique used in the "scaffolding algorithm" pictured in the figure here. $\endgroup$ Sep 16 '17 at 3:26
  • 4
    $\begingroup$ Just based on this problem, I would have noticed that 48*10=480 and that 436 is close to 480. So I would have actually started with larger numbers such as 48*9, 48*8 etc instead of starting at 48*1, 48*2 etc. And I've never seen BenjaminDickman's Scaffolding Algorithm taught in schools, but it makes sense and teaches kids that underestimating can still lead to the correct answer. $\endgroup$
    – ruferd
    Sep 18 '17 at 12:25
  • 2
    $\begingroup$ As a comment from someone who had to do this throughout school (and even while learning calculus, although this was on my own before college when calculators started becoming widely available), you almost never have to do more than 3 or 4 calculations. In this case, students would notice what @ruferd said or they'd notice that twice 48 is under 100, so you need at least 8 times 48 to get above 400, and you'd probably guess 9 because if 9 is too much then you know it has to be 8 (because you don't go higher than 9). $\endgroup$ Sep 18 '17 at 17:55

The awkwardness of "guessing" in the division algorithm is an artifact of the base-ten representation of numbers. If you represent in binary, then your only possible "guess" is 1.

In binary, your problem is to divide the six-bit number 110000 into the thirteen-bit number 1000100010000. Scanning the bits of the second number(the dividend) from left to right, there is no guessing needed when deciding where to put the leading 1 when forming the quotient above the dividend.

  • $\begingroup$ This is by far the best answer, because it takes all the B.S. out of the abstraction that is the base-10 number system! (although if base-10 is your first numbering system, you still have to use it to get your result back out again...) $\endgroup$
    – leeand00
    Sep 18 '17 at 16:21

Round. Check. Correct.

What we need is $436\div48$. Rounding to remove the last digit reduces the problem to $44\div5$. This gives us an educated guess of either $8$ or $9$.

From here we then multiply and check, either one. Take $8$ for an example. $48\times8=384$ and $436-384=52$.

By looking at what we have left, we know that we need only one more $48$ to go in, since our estimate was accurate enough to reduce the numbers down to things we can easily work with. By adding one more onto $8$, and subtracting $48$ from $52$, we get the first step:

\begin{align} \require{enclose} 9\hphantom{^4~8} \\ 48 \enclose{longdiv}{4~3~6^4~8} \end{align}

From which we can easily see the answer is $91$.


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