Update: For more background on the investigation below, I strongly encourage readers to visit Sol's wonderful blog wildaboutmath and, specifically, the post he wrote about the algebra behind multiplying 2- and 3-digit numbers. Further, he developed a video (mathcast) to demonstrate the method for 'mentally' mutiplying any two 3-digit numbers. Sol inspired me to develop this activity!

Target audience: Strong Algebra 1 or Algebra 2 students

Ok, I hope you enjoyed challenging yourself or your students to describe a step-by-step procedure for squaring three-digit numbers with a middle digit of zero. I believe many middle school students could devise a method using only operations on the digits, without the need to use algebraic representation or proof. I really believe that recording and organizing the necessary data and the search for patterns make that a valuable activity. Perhaps, even more importantly, the verbalization of the strategy or trick may be the most important benefit for our students! Finally, I strongly believe these kinds of investigations deepen student understanding of place-value and develop number sense.

But now we will move on to a much more significant challenge. I highly doubt that any of your students will be able to find the right combinations of digits to square any 3-digit number! One will still have to record the data and attempt to see a pattern, but, this time, a non-algebraic approach may be too formidable. I am fully aware that there are classical mental math methods for multiplying numbers (Vedic Maths and Trachtenberg to name a couple), but I didn't research those when writing this activity. I simply applied algebraic methods and looked for an algorithm that made sense to me. Thus my method may be similar to those others or not!

This challenge should be frustrating for some and many of you will question whether there is any benefit in looking for some strategy for squaring such numbers by some artificial-looking or convoluted combinations of digits. If the object here were to develop an easy mental math trick, you'd be right. BUT that is not my objective - the idea is to use algebra to motivate an alternate approach to multiplying 3-digit numbers. And, yes, there is some mental math involved, although, I suspect one will need to record intermediate results and do some paper and pencil arithmetic along the way (sorry, no calculator allowed other than to check your answer!). In the end, students should have a much deeper understanding of the meaning of digits, place value that is.

We begin with the algebra prerequisite for this problem, then move on to the real challenge!

READER/STUDENT CHALLENGE/ACTIVITY

(1) Show that

(p+q+r)^{2} =

p^{2} + q^{2} + r^{2} + 2pq + 2pr +2qr

Ok, here goes...

(2) Using the algebraic formula above, devise a method/strategy/algorithm/procedure/'trick' using combinations of digits to square ANY THREE-DIGIT NUMBER 'HTU'.

Note: Again, I reiterate, others have probably invented much cleverer methods for multiplication than this. The key here is to use the above algebraic representation!

In the end, you should be able to do an example 'mentally' like

463^{2} = 214,369 (or should I write it as 21-4-3-6-9).

Lots of 'carrying' here or 'regrouping' as we say these days! If I give you the intermediate step, it may give it away. If I don't, you may give up! Let's see what happens...

## Thursday, January 17, 2008

### A Much Harder Mental Math 'Trick'?? - Algebra May Not Be Optional!

Posted by Dave Marain at 8:13 AM

Labels: algebra 2, investigations, math tricks, mental math, number sense, place value

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## 5 comments:

Dave do you expect us to take

approach with the digit's place

using powers of 10 in the algebraic

calculation again?

The result would be very much

like in the previous challenge,

only that we have 6 instead of

3 terms to add (derived form your

formula above) .. or not?

Yes! And I know it's a real pain but at least one person enjoyed doing this -- me! Also, lots of regrouping to test your mental 'powers'!

One way to do it sequentially would be

f1=H^2

f2=f1*10+2*H*T

f3=f2*10+(T^2+2*H*U)

f4=f3*10+2*T*U

f5=f4*10+U^2

f5 is my final answer. I would like to lock it in:-)

TC

TC--

I accept your final answer!

I really like this 'nested' representation of a decimal number, which I discuss in more detail below. You've also set up a nice recursive description for the result, although I found the method that Sol used over at wildaboutmath is closer to the way I formulated the result. I will need to use both methods several times to see which is easier for me to keep track of mentally.

The nested form of a polynomial which is used to explain synthetic division is powerful and often overlooked in our curriculum. If

'ABCDE' represents any 5-digit number, then the actual value of this place value representation can be formulated as:

10(10(10(10A+B)+C)+D)+E. Try it!

Assuming I didn't mess up parentheses, this form is very useful and a great way for students to understand recursion. TC, what motivated you to use this? Do you see this as a more efficient algorithm?

When I add large numbers on paper, I tend to use the order from units to higher place values, as is typically done. However, when I add mentally, I find it easier to proceed from the largest to the smallest. I was just extending the same methodology to this case.

TC

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