Solving Basic Subtraction with Multiple Zeros

[Dav333]

how do i do:

8000
-7986
=

i have trouble with multiple zeros, thanks.

 

[arildno]

Well, you have 8 thousands.

You may write that like
$$8000=7000+900+90+10$$
(7 thousands+9 hundreds+9 tens+10 ones)
Thus, we have:
8000-7986=(7000-7000)+(900-900)+(90-80)+(10-6)=0+0+10+4=14

 

[Dav333]

thx. that's a bit over my head. :(

i was looking for the basic way kids do it, like this:

http://img18.imageshack.us/img18/3850/37640968.jpg

just when there are zeros together I get confused. thanks.

 

[arildno]

thx. that's a bit over my head. :(

s.

Well, develop your head a bit, then, and make an EFFORT to understand it.

Maths isn't a set of meaningless routines where what you are seeking is a Tutor who can say them in such a way that you memorize it.

 

[Mentallic]

Look at your example. You were trying to find out 8321-4031 but you turned 8321 (8 thousands +3 hundreds +2 tens +1 ones) into (8 thousands + 2 hundreds + 12 tens + 1 ones).

Now, can you try and extend the same idea to break up 8000 in such a way as to solve the subtraction problem? Arildno has the answer, and if you think about what he said, you should be able to apply that idea to the way kids do it.

 

[Bohrok]

The way I learned it years ago:
You move to the left until you get to the first nonzero number from which you can borrow. Cross that number out, subtract 1 from it, and write that number above it (cross out 8 and write 7 above it in your problem). Cross out all the zeros and write 9 above them except for the rightmost 0; that zero gets a 1 before it so it becomes a 10. Then you can subtract each column of numbers.