G

#### Guest

##### Guest

http://www.mikesmath.com/shortcuts.htm

Enjoy!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

G

http://www.mikesmath.com/shortcuts.htm

Enjoy!

Hmmm, I knew this stuff years ago ... so do most tutors here.

In fact, his explanations are a bit clumsy.

. . So I suspect he's an "amateur"

Subtract 50 from your number; call the difference \(\displaystyle d.\)

. . Example: \(\displaystyle 53^2\;\;\Rightarrow\;\;d \,=\, 3\)

Add \(\displaystyle d\) to \(\displaystyle 25.\)

. . \(\displaystyle 25\,+\,3\:=\:28\)

Append \(\displaystyle d^2\) (a two-digit number).

. . \(\displaystyle d^2\:=\:3^2\:=\:09\)

Therefore: \(\displaystyle \,53^2 \:=\:2809\)

Example: \(\displaystyle \,43^2\)

Subtract 50: \(\displaystyle \,d\:=\:43\,-\,50\:=\:-7\)

Add to 25: \(\displaystyle \,25\,-\,7\:=\:18\)

Append \(\displaystyle d^2:\;d^2\,=\,(-7)^2\,=\,49\;\;\Rightarrow\;\;43^2\:=\:1849\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

You need only Algebra 1 to verify this.

Let \(\displaystyle N \:=\:50\,+\,d\)

Then: \(\displaystyle N^2\:=\50\,+\,d)^2\:=\:2500\,+\,100d\,+\,d^2\)

Therefore: \(\displaystyle \,N^2\:=\:100(25\,+\,d)\,+\,d^2\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Jealous . . . me?

He gets credit for "inventing" these and gets on national TV.

Wasn't there anyone of the 20-20 TV staff who said,

. . "Hey, my son knows this stuff!"

I bet his next "discovery" is a proof that "1 = 2".

- Joined
- Dec 25, 2006

- Messages
- 28

Take any two-digit number, say 78. The closest 10 is 80, and the distance is 2. We do 80*76, which is 6080, and then add 2^2 to get 6084.

This method works in general because (

a+b)(a-b) = a^2 - b^2 ==> a^2 = (a+b)(a-b) + b^2, and it's not too hard to multiply a one-digit by a two-digit number in your head.

You can extend it to three digits, if you first imagine going to the near hundred, and then using the previous method for squaring the two-digit number as a difference.