Leapfrog Division IV

Published on Thursday, October 29, 2015 in , , ,

x divided by y fraction iconWhile I normally do my Leapfrog Division posts about a year apart, I though I'd wrap up this mental division series just 1 week after the previous entry.

In this post, you'll learn to mentally divide by numbers ending in 2!

STARTING POINTS: This is the most advanced technique of all of the Leapfrog Division posts, so you should be familiar with and practice the the previous techniques. Not only does this employ the basic ideas taught in the original Leapfrog Division post, but also the subtraction from 9 idea used in Leapfrog Division II, AND both the doubling and halfway-comparison concepts from Leapfrog Division III. If you comfortable with all of these concepts, then you're ready to move on to this version.

This version also introduces a new idea to the Leapfrog Division series: The stopping rule. In the previous versions, you could stop either when you realized the numbers were going to repeat, or when you didn't need any further precision. While the same is true here, in this version, you'll also need to stop when you get a quotient of 5, and a remainder of 0. You'll understand this better, including the exceptions, as you work your way through the technique.

THE TECHNIQUE: For our first example, we'll use 1732. As in the technique for dividing by numbers ending in 1, we will always start by reducing the numerator by 1, giving us 1632. Similar to the method for dividing by numbers ending in 8, we're also going to compare numbers to the half of the denominator. Anytime the numerator is greater than or equal to half of the original denominator, we'll reduce it by 1.

In the case of 32, half of that is 16, so we ask if the current numerator is greater than or equal to 16. In this example, it's currently 16 (exactly equal to half!), so we reduce by 1 more, giving us a division problem of 1522. As with all the Leapfrog Division techniques, we're now going to round the denominator to the nearest multiple of 10, and then divide it by 10. So, the problem becomes 153.

You shouldn't be surprised that we're going to divide this out using quotients and remainders. Starting with 152, we get:

  • 15 ÷ 3 = 5 (remainder 0)
Naturally, you write the quotient down right away, with a 0 and a decimal in front, as in 0.5. Ordinarily, the stopping rule might tell us to stop here. After all, we have a quotient of 5, and a remainder of 0. However, since 1732 (our original problem) isn't exactly 0.5, there's definitely more numbers to calculate, so we'll continue. We should still keep the stopping rule in mind for later, however.

From here, you're going to use the doubling idea as taught in Leapfrog Division III, in which you double numbers, but only keep the ones (units) digit. 5 doubles to 0, because 5 × 2 = 10, and we only keep the ones digit, which is 0. Next, as in Leapfrog Division II, you're going to subtract the quotient from 9. In this example, 9 - 0 = 9, so the new quotient is now 9. Leapfrogging the remainder, 0, to the front of the quotient, we know have 09, or simply 9.

Before dividing, we need to ask whether 9 is greater than or equal to 16. It isn't, so we don't decrease the number at this stage. After that question, only then do we do the division again:
  • 9 ÷ 3 = 3 (remainder 0)
We write the 3 down, giving us 0.53 so far. 3 doubled becomes 6, and 9 - 6 = 3. Leapfrogging the remainder of 0 in front, we have 03, or just 3. Is 3 greater than or equal to 16? No, so there's no decrease this time, either. The next division problem yields:
  • 3 ÷ 3 = 1 (remainder 0)
So far, our mental work has given us an answer of 0.531. Repeat once more. Double the 1 to get 2, subtract 9 - 2 to get 7, and put the remainder 0 in front of the 7, giving us a new divisor of 07. Is 7 greater than or equal to 16? No, so there's no decrease. Moving on to the next division:
  • 7 ÷ 3 = 2 (remainder 1)
Now we have 0.5312. 2 doubled becomes 4, and 9 - 4 = 5. Leapfrogging the 1 in front gives us 15. Is 15 greater than or equal to 16? No, so there's no decrease. Obviously, we can move on to the next division:
  • 15 ÷ 3 = 5 (remainder 0)
At this point, with 0.53125 as our current answer, you'll note we have a quotient of 5 and a remainder of 0. This means that the stopping rule kicks in. So, our final result is 0.53125, which is exactly what 1732 equals!

Now that you understand the steps, let's work out 1922 as a second example. We start by reducing 19 by 1, which is ALWAYS the first step, giving us 1822. Half of 22 is 11, and 18 is greater than 11, so we decrease it by 1 again, leaving us with 1722. Rounding the denominator down and dividing by 10, our starting problem should be 172. We start there, and work through the problem this way:
  • 17 ÷ 2 = 8 (remainder 1)
    (8 doubles to 6, 9-6=3, 1 makes it 13, which is MORE than 11, so 13 - 1 = 12.)
  • 12 ÷ 2 = 6 (remainder 0)
    (6 doubles to 2, 9-2=7, 0 makes it 07, which is less than 11.)
  • 7 ÷ 2 = 3 (remainder 1)
    (3 doubles to 6, 9-6=3, 1 makes it 13, which is MORE than 11, so 13 - 1 = 12.)
Since we're already back to dividing by 12, you can see that this is going to repeat. Writing down just the quotients, we get the correct answer of 1922 ≈ 0.863!

TIPS: Yes, this has more steps than any of the other approaches taught in the Leapfrog Division series, and it's not difficult to confuse the steps of the various versions. The solution, as always, is practice, practice, practice!

You may have noticed that I referred to this as the last post in the Leapfrog Division series. Why is that? Because using the 4 different techniques I've taught, you can actually handle dividing by most numbers with just a little adjustment. How do you handle numbers ending in...
So, every number except those ending in 5 or 0 are covered. That's good incentive to practice these techniques, as you can amaze many people with the ability to handle almost every division problem thrown at you. I hope you've found this series enjoyable and useful.

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