## Tuesday, January 15, 2008

### 508^2 = 258,064: Invent your own Mental Math 'Trick' and Prove It!

UPDATE: READ THE COMMENTS FOR A DISCUSSION OF NOT ONLY A POSSIBLE SOLUTION BUT ALSO THE PEDAGOGICAL IMPLICATIONS OF SUCH AN ACTIVITY.

What child (ok, adults too!) is not mesmerized by a magician. The Hindu-Arabic place value system allows for endless possibilities for
mathemagic, aka, mental math strategies. Adults and children alike are fascinated by these as is apparent from the re-discovery of Vedic Maths. There are several excellent resources for this. You may want to check out The Vedic Maths Forum India Blog, which includes an interesting video showing youngsters demonstrating one of these methods in a dance routine! It's also no wonder that Sol's Impress your friends with mental Math tricks post over at wildaboutmath went viral! Sol discusses many of these wonderful mental math strategies and the response has been overwhelming.

I've personally always wanted to understand the basis for many of these mental math 'tricks' that have been around for centuries. Since a significant part of teaching is performing, math educators are often interested in any mathematical sleight of hand that provokes wonder in the child or older student. We would hope the student would want to know the WHY behind the trick but most are more interested in just performing the feat. What child doesn't come home and want to challenge their parents to square 45 in their heads or multiply 68 by 62 in 5 seconds or less! Hey, anything that turns kids on to mathematics is alright in my book. However, the purpose of this post is to have the reader discover a mental math method and use algebra to delve beneath the surface. Magic? Perhaps that is what teaching is all about...

Study the following:
1012 = 10201
4012 = 160801
5082 = 258064

3062 = 93636
9092 = 826281

1. From studying the above examples, discover a mental math strategy which would enable you to square any 3-digit number whose middle digit is zero. Write the precise steps of the method. Be careful here - make sure your method works for all 5 of the above examples! Note: That last one is a bit harder to do mentally but give it your best.

2. Demonstrate (perform!) your method by having someone challenge you to square such a number! You need to be able to perform the trick in less than 5 seconds (ok, maybe 10 for old people like me!). Repeat your amazing performance at least 5 times!
Note: You may need to practice this for awhile before going on American MathIdol!

3. Ok, now PROVE IT ALGEBRAICALLY!
(Gee, that takes all the fun out of it.)

(i) Did you ever notice that all of these multiplication tricks actually require that the child knows the basic facts cold! Perhaps, these tricks could even serve as motivation to learn them!

(ii) When I discovered this trick, I was naturally excited and of course wanted to share it with my wife who has a droll way of putting my geekiness in perspective. After she picked a random 3-digit number and I did the math correctly, she seemed unimpressed and replied, "But can you cook a chicken?" I was tempted to suggest that I could square a chicken if it had a hole in the middle, but I knew she would not be amused....

Sol Lederman said...

Dave,

This is a great exploration. I won't give the algebra away but I do see how it works!

You and your readers might enjoy my recent posts on Vedic multiplication and bases. It's quite fascinating how you can simplify a number of multiplication problems with a very interesting approach.

MathFr3d said...

I'll try to give a proof first and the the technique in words:

Let n = a*10^2 + b be the number to be squared.
(such that it has the form "a0b")

Then n^2 equals:

n*n =
(a*10^2 + b)*(a*10^2 + b) =
(a*10^2)*(a*10^2 + b) + b*(a*10^2 + b) =
(a^2)*10^4 + 2ab*10^2 + b^2

In words this means:

1. The last two digits of n^2 hold the value of b^2.
2. The next three digits (**) contain the product of 2ab.
3. Then add a^2 to the result starting with the 5th digit.

(**) if n is smaller than 317 (about square root of 10^6)
then n^2 is a 5 digits number, otherwise it has 6 digits.

Remembering this makes step 3 easier, because we dont have
to add a^2 to the first digit of 2ab in the 4th place, we
can simply put a^2 into the 4th and 5th place of n^2.

Ok now I have to practice this and
go find someone to impress ;)

MathFred said...

I think I made a small mistake above with the hint when n<317.

The correct way to recognize when
the highest digit of 2ab has to be
added to a^2 is when 2*ab > 99 or
a*b>49.

The boundry n<317 is only a hint when
the n^2 has 6 and not 5 digits.

Dave Marain said...

Thanks, Sol. I hope you know that you inspired me to write this student activity!

mathfr3d--
Excelent analysis. The issue of the middle digits coming from the 2ab*10^2 term is probably the only part that would give students difficulty. I'm guessing that most students would avoid the technicalities by saying something like:

Double the hundreds' digit of the original number, n, then multiply it by the units' digit (uh, of course, most would say last digit or ones' place). If the result has 2 digits, insert those in the middle. If the result has 3 digits, say 162, for example, then insert 62 in the middle and carry the '1' over to the left (your step 3).

By the way, the only possible 'carry' would be 1, as in squaring 909. This 'carry' would occur, as you said, when 2ab>99 or ab is greater than or equal to 50. This occurs only in the following 6 cases:
a=6,b=9
a=7,b=8; a=7,b=9;
a=8,b=7; a=8,b=8; a=8,b=9
a=9,b=6,7,8 or 9
My guess is that students are more likely to list these cases than give the algebraic condition but that probably depends on their sophistication.

The other piece that has to be mentioned is if 2ab or b^2 are single digits as would occur, say, if n = 201 for example. In this case the middle digits would be '04' since there must be exactly two digits in the 'middle', namely the 3rd and 4th digits. The same thinking applies to the first 2 digits which arise from b^2. There are again TWO places to fill and in our example this would be '01'.
Thus, 201^2 = 4 04 01, inserting spaces for emphasis.

Dave Marain said...