In 2013, I posted about Leapfrog Division, which was A.C. Aitken's approach for mental division by numbers ending in 9. In 2014, I built on this method with Leapfrog Division II, an approach for mentally dividing by numbers ending in 1.

It's 2015, so it's time for another update to the Leapfrog Division technique. This time, you'll learn the method for mental division by numbers ending in 8!

**STARTING POINTS:** You'll want to be very familiar with the process of dividing numbers by 9, as taught back in the Leapfrog Division post. There are a few extra steps in this version, as compared to the original version, so being well versed in the original is imperative. You may also find it helpful to have practiced the technique for dividing by numbers ending in 1, as taught in Leapfrog Division II, but that's not as essential to this approach.

When teaching this technique, I'm going to be referring to doubling a given number, but they're doubled in a special way. As used in this technique, you double the number, but only keep the ones (units) digit. Ordinarily, you would double 5 to get 10, but here you only need to remember the 0. In a similar manner, 6 doubled will give you 2 (12, with the tens digit dropped), 7 doubled will give you 4, 8 doubled will give you 6, 9 doubled will give you 8, and 0 doubled will give you 0.

**THE TECHNIQUE:** As our very first working example, we'll work out the decimal equivalent of ^{13}⁄_{18}. Just as before, you're going to start by rounding up the denominator (the bottom number) to the nearest multiple of 10, and then drop the 0. In our example, that means that our fraction gets changed to ^{13}⁄_{20}, and dropping the 0 from the denominator changes this to ^{13}⁄_{2}.

We're going to ask a question here which will be asked over and over again, and this question will help give us the correct total. Is our current numerator greater than or equal to half of the *original* (before rounding) denominator? If so, we *MUST* add 1 to it. For example, half of our original example denominator, 18, comes to 9. So, we're going to be asking at the start, and several points afterward, whether our current numerator is equal to or greater than 9. To start, we realize that 13 is equal to or greater than 9, so we add 1, giving us ^{14}⁄_{2} as our actual first problem to solve.

We're going to work this out in a similar manner as before, solving this division problem with a quotient and a remainder. As you go, you're going to write down quotients as you go, and keep remainders in your head. Our first division yields:

- 14 ÷ 2 = 7 (remainder 0)

How do we take the next step? First, the quotient (the 7 in our example answer above) must be doubled. Don't forget that we drop the tens digit when doubling! So, we double 7 to get 14, and drop the tens digit, leaving us with 4. The remainder (0, in the problem above) then “leapfrogs” to the front of the 4, giving us 04 as our new numerator (which is just equal to 4, of course).

We ask ourselves one more time, is our current numerator (4) greater than or equal to 9? In this case, 4 isn't greater than 9, so we don't add 1. After that, we divide by 2 again to get:

- 4 ÷ 2 = 2 (remainder 0)

2 (the quotient) gets doubled again, to make 4, and the remainder of 0 leapfrogs in front to give us 04, or 4, once again. Is this new 4 greater than or equal to 9? No, so we leave it alone. Dividing by 2 one more time yields:

- 4 ÷ 2 = 2 (remainder 0)

^{13}⁄

_{18}is indeed 0.722..., with the 2 repeating endlessly.

Just to help lock in the technique, let's try and work out the decimal equivalent of

^{9}⁄

_{28}. We have to remember to keep asking ourselves about half of the original denominator, which is 14 this time. Is 9 equal to or greater than 14? No, so we won't add 1 at this point. The denominator gets rounded up to 30, and we drop the 0 to leave us with a starting calculation of

^{9}⁄

_{3}:

- 9 ÷ 3 = 3 (remainder 0)

(3 doubles to 6, 0 in front makes 06, which is less than 14.) - 6 ÷ 3 = 2 (remainder 0)

(2 doubles to 4, 0 in front makes 04, which is less than 14.) - 4 ÷ 3 = 1 (remainder 1)

(1 doubles to 2, 1 in front makes 12, which is less than 14.) - 12 ÷ 3 = 4 (remainder 0)

(4 doubles to 8, 0 in front makes 08, which is less than 14.) - 8 ÷ 3 = 2 (remainder 2)

(2 doubles to 4, 2 in front makes 24, which is*MORE*than 14, so 24 + 1 = 25.) - 25 ÷ 3 = 8 (remainder 1)

(8 doubles to 6, 1 in front makes 16, which is*MORE*than 14, so 16 + 1 = 17.) - 17 ÷ 3 = 5 (remainder 2)

(5 doubles to 0, 2 in front makes 20, which is*MORE*than 14, so 20 + 1 = 21.) - 21 ÷ 3 = 7 (remainder 0)

(7 doubles to 4, 0 in front makes 04, which is less than 14.) - 4 ÷ 3 = 1 (remainder 1)

(1 doubles to 2, 1 in front makes 12, which is less than 14.)

^{9}⁄

_{28}is equal to 0.32142857..., with the 142857 repeating. Actually, if you know the 142857 pattern from knowing your 7ths, and you realize that 28 is a multiple of 7, you should realize that you'll eventually run into the 142857 pattern from there.

**TIPS:**As always, the biggest tip is practice, practice, practice! Once you can divide by numbers ending in 8, you should also realize that you can divide by numbers ending in 4. If you want to divide by a number ending in 4, just double both numbers in the problem. If you need to work out

^{17}⁄

_{24}, for example, just double both numbers, resulting in

^{34}⁄

_{48}, and work the problem out from there, as described in the technique section. As a matter of fact, you'll get a great deal of practice if you work out

^{34}⁄

_{48}on your own, right now.

## 2 Response to Leapfrog Division III

I'm so happy this site is back. I check here very often to see if it has been updated and I am very happy to see that it was. It's funny because I have been playing with division lately using the techniques from the mental calculator's handbook and dead reckoning. I wasn't aware of this trick. I usually factor the denominator and then divide twice but this is better. Thank you very much

It's good to see you're back! I was getting a little concerned. I have no way to contact you except this site or I would have tried. Just wanted to make sure you know that this website is great and would be greatly missed if it were to go away. Thanks again for all the work you do for our entertainment.

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