In Part 1, we discussed how to multiply two 2-digit numbers together. For Part 2, we will be discussing how to multiply two 3-digit numbers. Once we learn the method, we will be able to expand that method to apply to numbers of any size. Like before, I will be working from left to right (mainly because of how wordpress formats).
Multiplication of two 3-Digit Numbers
Usually I try to make sure the first example doesn’t contain any ‘carry-overs’, since they tend to complicate things a little. However, in this case there will be carry-overs. The reason being, since we are multiplying two 3-Digit numbers I would like to use 6 unique digits so you can keep track of which numbers I’m multiplying. Once you understand the pattern, then we can try other examples which contain the same digit multiple times (i.e. 343 x 122, 567 x 657, etc….things like that).
For this example, we’re going to multiply 123 x 456 (see 6 unique digits).
1) Write out the two numbers in the following way (or visualize it…this may take a little work keeping track of all the numbers).
1 2 3
4 5 6
————-
2) Since I’ll be working from left to right, the first thing we will do is multiply the left-hand side vertically. 1 x 4 = 4….write ’4′ down.
1 2 3
4 5 6
————-
4
3) Next we will cross multiply and add. (1 x 5) + (4 x 2) = 13. Since we can only keep one digit, we keep the ’3′ and carry the ’1′. Now the first term becomes ’5′…(4+1).
1 2 3
4 5 6
————-
5 3
4) This is the new step. We are going to cross multiply (1 x 6) and (4 x 3), and multiply the middle term vertically (2 x 5). Add these three numbers together…6 + 12 + 10 = 28. We put down the ’8′, and carry the ’2′ over. The second term now becomes ’5′…(3 + 2).
1 2 3
4 5 6
————-
5 5 8
5) Now we continue with the cross multiplication, this time multiplying (2 x 6) and (3 x 5). Add these two numbers together…12 + 15 = 27. Now lot’s of carrying-over is about to happen. We write down the ’7′ as the fourth term, and carry the ’2′. We add this ’2′ to our third term (’8′)…but this now equals ’10′. We write down ’0′ in the third place, and carry the ’1′. This makes our second term ’6′ now … (5 + 1). I hope that didn’t confuse you…if it did let me know in the comments and I’ll try to clear it up.
1 2 3
4 5 6
————-
5 6 0 7
6) And finally we multiply the right-hand side vertically. (3 x 6) = 18. We write down the ’8′, and carry the ’1′. Add the ’1′ to our fourth term (’7′) to give us ’8′, write that down in the fourth spot.
1 2 3
4 5 6
————-
5 6 0 8 8
Answer: 123 x 456 = 56,088
Algebraic Proof: Multiplication of two 3-Digit Numbers
We will represent our two numbers as the terms (ax^2 + bx + c) and (dx^2 + ex + f), where x = 10, and the other variables are numbers between 1 and 9.
For this first step, the right-hand side of the equation continues on the second line…formatting errors are preventing me from writing it neater (Sorry!). So please bear with me, while I attempt to make it somewhat readable.
(ax^2 + bx + c) * (dx^2 + ex + f) = (ad)*(x^4) + (ae)*(x^3) + (af)*(x^2) + (bd)*(x^3) + (be)*(x^2) + (bf)(x) + (cd)*(x^2) + (ce)(x) + cf
(ax^2 + bx + c) * (dx^2 + ex + f) = (ad)(x^4) + (ae + bd)(x^3) + (af + be + cd)(x^2) + (bf + ce)(x) + cf
It ends up looking like this…
a b c
d e f
—————
(ad)(x^4) + (ae + bd)(x^3) + (af + be + cd)(x^2) + (bf + ce)(x) + cf
Or if all those exponents and extra addition signs make things confusing for you…here’s another way to look at it. Split the answer up into 5 different parts, which are show below separated by ‘|’ sign’s.
a b c
d e f
—————
(ad) | (ae + bd) | (af + be + cd) | (bf + ce) | (cf)
To figure out how many parts its going to ultimately need to be split up into you can just use this simple formula…
n + (n-1)
where ‘n’ equals the number of digits of the larger number. For instance 123 has ’3′ digits…so 3 + (3-1) = 3 + 2 = 5. Say your multiplying 456 x 3…how many digits? Three again…it works the same as before, except you write 3 as 003. Applying the steps above, you end up with:
4 5 6
0 0 3
———–
1 3 6 8
See if you can do some of these practice problems…some contain more than three digits.
Practice Problems:
1. 525 x 122
2. 2451 x 32
3. 621 x 9
4. 243 x 133
5. 3123 x 4421
I’m sorry about how messy the algebraic proof turned out to be. Things will get a lot neater once my other laptop is up and running again. Once that happens, I’ll write up the proofs along with some more examples as a pdf file that you can save.
By the way, which way is easier for you to read …the proof with the exponents and ‘+’ signs, or separating the parts by using “|”‘s?
I’m having a little trouble because you didn’t show examples (like numbers two and three above in the practice problems) where the two numbers being multiplied have different numbers of digits.
But I’m making progress. Back soon.
Eileen
Dedicated Elementary Teacher Overseas
elementaryteacher.wordpress.com
Sorry about that! I didn’t even notice that I skipped out on a bunch of examples (must of been sleepy haha). I’ve posted an example below; if you would like me to post more examples, I can make a separate post this afternoon.
Example: 326 x 4
In order to do this, we can add two zeros in front of the number ’4′ to make it a three-digit number. So our problem then becomes 326 x 004.
3 2 6
0 0 4
————–
1) The first thing we do is multiply the left-hand
column vertically. (3 x 0) = 0
3 2 6
0 0 4
————–
0
2) Next we “cross multiply and add” the first two
columns (corresponds to Step #3 in the post). (3 x 0) + (0 x 2) = 0.
3 2 6
0 0 4
————–
0 0
3) Now we follow Step #4 in the post, which tells
us to “cross multiply and add” the outer columns…while vertically
adding the middle column. (3 x 4) + (0 x 6) + (2 x 0) = 12. Since
we can only have one-digit, we write down the ’2′ and carry the ’1′
to the left.
3 2 6
0 0 4
————–
0 1 2
4) Now we “cross multiply and add” the last two
columns. (2 x 4) + (0 x 6) = 8.
3 2 6
0 0 4
————–
0 1 2 8
5) Now we multiply the last column vertically.
(6 x 4) = 24. We write down the ’4′, and carry the ’2′ to the
left (I’ll just write it below the ’8′ for now instead of
combining them). Note: Ignore the “…..” it’s just there for
formatting.
3 2 6
0 0 4
————–
0 1 2 8 4
………2
6) Adding them together (8 + 2) gives us ’10′ for
that column. We write down the ’0′, and carry the ’1′ to the left.
3 2 6
0 0 4
————–
0 1 2 0 4
……1
7) Add those together to get ’3′ for that column.
3 2 6
0 0 4
————–
0 1 3 0 4
Answer: 326 x 4 = 1304
I’m still trying to figure out how everything works in wordpress…how do you get it so that it only shows a summary of each post on the front page?
Please, I worked through all your examples, but was unable to figure out the procedure for multiplying the two four-digit numbers. Could you please give me a full example of that, too? (Sorry!)
Best regards,
Eileen
Click on your July Archives (button on your front page) and it will show each post with a brief summary.
Eileen
Here you go, it’s in a pdf file. Once you get past two 3-digit numbers it get’s hard to explain with words, so I tried to do it visually for you.
4-digit Multiplication (pdf)
Thank you. Now I could follow this example. I see there is something you forgot to say in your original instructions. For the examples you gave first (above) we are multiplying from left to right (starting with the larger place values, and finishing with the ones’ place). In the PDF example, they are STARTING with the ONES’ place, and going in the traditional direction toward the larger place values. When I first tried to do number five above, I was still trying to work from the left, moving right. I was able to follow the PDF example, it was very clear.
However, I have to say that it seems really neat in multiplication of two digit numbers, followable in three-digit numbers, but a bit hard to keep track of the changing place values in answers (and wouldn’t be possible for most people mentally). However, by the time you are getting to four digits, I personally think the traditional method is much easier in terms of keeping track of place values. What do you think? I’m not seeing the advantage of using Vedic multiplication in larger numbers; I am only seeing the advantage when multiplying smaller numbers.
Comments?
Eileen
Dedicated Elementary Teacher Overseas (in the Middle East)
elementaryteacher.wordpress.com
Actually, all of these multiplications can be done from either side; left to right OR right to left. The reason I usually do it from left to right is that when I do it that way I am able to say off the answer as I’m computing it.
I agree that it is hard once it gets to 4-digits (I still have a hard time mentally since I haven’t practiced it too much), however it is doable with some practice. What I suggest is first is to continue doing it with 2-digit numbers until you have it down solid. Then once that happens move up to 3-digits, keep practicing it until you can learn to memorize the digits. It’s hard at first but after a little while it becomes easier.
Also, the multiplications will become ALOT easier once you begin using viniculumn numbers. They basically allow you to only have to know the multiplication tables up to 5 x 5. Using them you only use the numbers 0-5, since you convert the higher numbers (6-9) into the lower numbers (sounds confusing I know, but its easy once you see it…just haven’t found a font yet that has “bars” over the numbers). Once you’ve converted the numbers to viniculum numbers, your twice as likely to find yourself multiplying by 0 & 1, which makes it really easy.
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