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I just realized some simple math...

Chaotic42

Lifer
If you multiply a number ending in a certain digit by a number ending in that same digit, the last digit of the product will be the same as it would if you multiplied a number ending in (10-[digit]) by another number ending in (10-[digit].
Example:

9*9=81
(10-9)*(10-9)=1

8*8=64
(10-8)*(10-8)=4

7*7=49
(10-7)*(10-7)=9

See what I mean? I can't believe that I hadn't noticed this before.
 
Look up "casting out nines". It is a fun little trick you can use to check your math. Instead of busting out a calculator to balance the check book, flex your brain muscles, do the math by hand, and use a technique like casting out nines to check your work.

R
 
Originally posted by: rgwalt
Look up "casting out nines". It is a fun little trick you can use to check your math. Instead of busting out a calculator to balance the check book, flex your brain muscles, do the math by hand, and use a technique like casting out nines to check your work.

R

Heh, I picked up a speed math book and that's what got me thinking about numbers on my trip home from work. Here's a trick from the book:

For multiplying numbers close to 100 (A*B), subtract B from 100 and then subtract that number from A. Then multiply (100-A) * (100-B). Add that to 100*A and that's A*B.

99x99

99-1=98
(100-99) * (100-99) = 1
98*100=9800
9800+1=9801=99*99

Pretty neat.
 
Here's what I figured out by myself in grade 3 for the 9 multiplications

9 * x = y
9 * 10 - x = y

Use it to this day 😉
 
Originally posted by: bootymac
Here's what I figured out by myself in grade 3 for the 9 multiplications

9 * x = y
9 * 10 - x = y

Use it to this day 😉

9*4=36
9*10-4=86

It's late, am I doing something wrong?
 
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