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The algebra of cross-multiplication

A few people were underwhelmed with my video on how to multiply together a pair of 2-digit numbers without writing down partial results. They didn't see a time savings. Fair enough, although not needing to write anything but the three (or four) digits of the answer can be a nice time savings and a source of smugness for many.

Next week I'm going to produce some more mathcasts. The first one will show how to multiply together a pair of 3-digit numbers using the cross multiplication approach. I think that will impress more people.

There's a general approach to these multiplications I want to teach you. If you understand the algebra behind this approach you can derive the steps for multiplying together numbers containing any number of digits. However, I'll warn you that beyond 5 digits this approach, and any other approach that works for any numbers and not just special cases, will really work your mental muscles.

Let's look at the 2-digit case where we want to multiply ab by cd, where a, b, c, and d are the digits of the two numbers. ab really means 10a+b and cd really means 10c+d. In other words, when we write the number 23 in our familiar base 10 we mean 2 tens and 3 ones, or 2*10+3.

We calculate ab x cd by doing the algebra:

(10a+b)x(10c+d) = 100ac + 10ad + 10bc + bd

We can group by powers of 10 and we get:

(10a+b)x(10c+d) = 100ac + 10(ad+bc) + bd

We see that the ones digit is bd, which may produce a carry. bd is just the product of the digits in the right column, just like in the video.

The tens digit is ad+bc, which is the cross-product illustrated in the video for the middle steps of the calculation. Again, carries may be involved.

The final part of the calculation, the hundreds place, is ac, plus any carry. That's the product of the digits in the left-most column.

Let's look at the 3-digit case, abc x def:

(100a+10b+c)x(100d+10e+f) = 10000ad + 1000ae + 100af + 1000bd + 100be + 10bf + 100dc + 10ec + cf

Let's regroup the terms by powers of ten. We get:

10000ad + 1000(ae+bd) + 100(af+be+dc) + 10(bf+ec) + cf

I'm going to save the visualization for the next video but, using the algebra we just did, you can see what pairs of numbers you have to multiply and add to get the digits of the product.

Hopefully you can see how to extend this approach to multiply together numbers with 4 digits and beyond. If you're able to keep a running total of all those products then you won't need to write down any partial products. The biggest trick here is to see the pattern behind what digits to multiply together. If you derive the recipes for 4, 5, and 6 digit cross multiplication, cumbersome as this may be, you'll start to see a general pattern.

Comments (5) Trackbacks (6)
  1. how do you find the 3-digit numbers and the product in a multiplication of 2 , 3 digit numbers ,if the numbers are replaced by alphabets ??
    ex: B H I * C A F
    B A J C
    D D E E
    J G H D
    J A B G G C

    HERE each number is replaced by an alphabet and no two alphabets represent the same number .

  2. it is so awsome and it is so easy i nt understand why i dint do goo on the exam i need more lol

  3. please give me full method of doing this alphabet multiplication.how those alphabets will be converted into digits from 0-9 to get desired result..

  4. can any body provide me a complete solution of this problem

  5. please give the solution with full explanation…..

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