Math Shortcuts from Mike Byster

Hello, AirForceOne!

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"


To square numbers "near 50"

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

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

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Jealous . . . me?

Of course, I am!

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".

 
I would only comment that most of his shortcuts look extremely specific, and not very useful unless you happen to fit into one of the categories. In terms of squaring 2 or 3 digit numbers, Arthur Benjamin demonstrated one that works for them all.
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.
 
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