Russian Peasant Non-Standard Algorithm Demonstration

This video was created as a means of academic evaluation for my Educ 4173 Elementary Mathematics class at Acadia University. I will demonstrate a non-standard algorithm called the Russian Peasant method in multiplication.

**Note: This video is an edit of a previous video, still viewable on YouTube, but includes audio narration and insertion of extra footage in which the "partial products" page is not removed during transfer of the four numbers to a new page. This allows for logical transition between pages and better understanding of the next step in the process.

Category:

Education

Tags:

• Russian
• Peasant
• Multiplication
• Non-Standard
• Algorithm
• Mathematics

[midinerd]

very interesting any instance were the measurement is corrupt, it's just ignored and sums the rest... awesome

[MissJHines]

I'm not quite sure what you mean by ignored or corrupt. The same algorithm is followed consistently and nothing is overlooked while computing the formula. Thanks for stopping by.

[midinerd]

I didn't even type "where" correctly, and I didn't state what I really meant.

I wanted to say if some certain side wasn't prime, to ignore it. I guess that is close to 'even,' and further - both even. I feel like the reason behind it is there but I have to find it...

I think maybe it means there are unique instances of proportional quantities that should be calculated differently? I have no idea but this video is very inspirational.

[MissJHines]

ok, I see what you are saying. Yes and no to your question. one side or the other is prime, you cancel out that line. If both numbers are even, you keep them for the final outcome. Did that make sense?

[MissJHines]

that's why it's an alternate algorithm, the idea was to show alternate means to solve an equation, not find the easiest :)

Thank you for visiting and for posting a comment

[Freakshow05]

this works only if the first number of the multiplication is uneven.

Try 88x12, or 66x35.

it doesn't work......besides, you would be faster with the original multiplication method......although i think it's cool to find a new way (so, no offense)...but unfortunately it doesn't work for every multiplication...

[MissJHines]

to illustrate

88 12----this row is crossed out

44 24----this row is crossed out

22 48----this row is crossed out

11 96----this row remains

5 192----this row remains

2 384----this row is crossed out

1 768----this row remains

Eliminating all rows where both numbers are even (including the 88x12 row), you get 768+192+96= 1056 and to proof it,88x12=1056

The algorithm does work, you just have to know how to work it. If the issue is the order of the numbers, switch them

[returnoftheramble]

@vagelis17y this was designed to make people slow. I'm writing a book to confuse even the best of students.

[returnoftheramble]

I can do these all day long. You can find the next square if you follow this pattern

First take the square that you know

3^2=9

Take the square root of 9

which is 3

Then add the next number, 4,

which is 3+4

Then add that total, 7, to the product of step 1, which is 9+7

The total is 16 and that's the answer to 4^2

[vagelis17y]

simple=easier

[MissJHines]

The idea isn't to have an expedient method of multiplication, but an alternative algorithm, which was the purpose of the assignment. Of course, if you chose the traditional computation method, you'd find your product much faster. That simply wasn't the purpose of the method at all. The 88x12 does work, you are trying to add 12 to the list of addends, but must remember that 88 and 12 are both even numbers, so they get crossed out leaving you 96+192+768 which does equal 1056, the product.