04 Mar The Math behind Xinwan’s Trick
In a previous post, I wrote a bit about a mental math method that I refer to as Xinwan’s trick. It involves multiplying two-digit numbers times eleven by splitting the digits and placing the sums between them. For instance, to multiply 62 x 11, you simply split the 6 and 2, like this: 6___2, then add them together and place the 8 in the middle. So, the answer is 682.
It’s a pretty great trick, and the students really love using it. I ended my earlier explanation of it by mentioning my guess that one of the reasons the students really embrace it, is that it has a certain magical appeal – it’s surprising that it works. The thing about math though, is that if something consistently works, there’s a reason why it does so. So why does Xinwan’s trick work? It has to do with an alternate method of multiplication that has fallen out of practice, but was once, as I’ve read, quite common.
Here’s how you do it:
- First, multiply the tens digits, then the ones digits, then cross multiply and add, placing the sum inbetween the products of the tens and the ones.
- That sounds confusing, so let’s try an example: 32 x 43.
- Begin by multiplying the tens, 3 x 4 = 12.
- Next, multiply the ones, 2 x 3 = 6.
- Then, cross multiply and add, (3 x 3) + (2 x 4) = 17
- Put it all together: 12_17_6.
- Finally carry the one from the tens column over to the hundreds, because 17 tens is really 170.
- Your answer is 1,376.
It’s not a bad method, with a little practice it’s actually a pretty good option for doing some two-digit by two-digit multiplication. The only downside is that most problems will require at least some carrying – which can get confusing, and lead to errors.
So, what does this have to do with Xinwan’s trick? Well, it turns out that Xinwan’s trick is really just a special case of this multiplication method. “One” of course is the multiplicative identity. That simply means that any number times one, equals itself. So, let’s try this method with a number times 11.
Let’s multiply 53 x 11.
- First multiply the tens, and then the ones: 5 x 1 = 5, 3 x 1 = 3. Notice that because multiplying a number times one returns the original number, we don’t really need to bother doing it at all.
- Next, cross multiply and add. Again, since we’re multiplying times one, we don’t even need to do that step, we can just skip straight to the adding part: 5 + 3 = 8.
- Put it in the middle and Boom! 538!
So, when using this method to multiply times 11, we can skip all of the multiplication, because it only involves numbers being multiplied by 1. All we need, is to add the numbers and place them in the tens column between the original digits. The most work we might have to do, is to carry a one – but that can only ever be from the tens to the hundreds column, never from the ones to the tens, and it is only necessary in half (45 out of 90, do you see why?)* of the potential problems!
One last thing – you might be wondering – why does this method of multiplication work at all? Lets pull it apart and take a look. Let’s use our original example again, 32 x 43. We could rewrite this as (30 + 2)(40 + 3). Using the distributive property (some of you might have learned this as FOIL: First, Outer, Inner, Last), we can multiply the first terms (30 x 40) and add that to the products of the outer terms (30 x 3), the inner terms (2 x 40) and the last terms (2 x 6). This gives us 1,200 + (90 + 80) + 6 = 1,200 + 170 + 6 = 1376. This is essentially the same as the method outlined in the example above, just with the zeroes explicitly defining the tens and hundreds columns.
So that’s it. It turns out it wasn’t magic at all, although I hope that knowing that it was really just math all along, makes it (and math!) even more fascinating for you than it was before – it does for me!
*There are 90 instances of two digit numbers that can be multiplied by 11. Xinwan’s trick of course works with single-digit numbers, you just need to consider the tens digit to equal zero, but you’ll obviously never have to carry. The two digit numbers range from 10 – 99. Of the numbers from 10 to 19, only 19 has digits that sum to a two-digit number. Of the numbers from 20 to 29, only 28 and 29 have two-digit digit sums. Likewise, there are three numbers in the thirties, four in the forties and so on, up to nine in the nineties that have double digit digit sums, and so, would require carrying. This totals to 45, we could write it out like this, 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45. Do you know how to quickly find the sum of a series like this? Notice that the first and last terms add to ten, as do the second and the next to last terms, as well as the third and third from last, and so on. So, you have four tens, plus an unmatched five, for a total of 45 – pretty neat, huh?!