The Nine's Method

[Barnpreacher]

Have any of you logical, brilliant minds ever heard of this? Someone in my church showed it to me tonight and asked if I could help her figure it out so that she can receive bonus points in her college course.

Create a subtraction problem (have a subtrahend with three digits and a minuend with three digits, but don't subtract it yet). Subtract the subtrahend from 999. Then add the difference to the original minuend. When you get the sum cross out the number in the thousands place and add the number that you cross out. That gives you the answer to the original subtraction problem.

Example problem: 727 - 397 (that's the original subtrahend and minuend)

999 - 397 = 602

602 + 727 = 1329 (cross out the 1)

329 + 1 = 330 (the answer to your original problem)



How or why does it work everytime? Thanks!

 

[panta dokimazete]

It looks to me that you are basically solving your original problem, just more roundabout by throwing in the value of 1000 then eliminating it.

 

[panta dokimazete]

here is a like model using the value of 100 instead of 1000

87 - 57

99 - 57 = 42

42 + 87 = 129 (cross out the 1) [which is actually subtracting 100]

29 + 1 = 30

I'll see if I can't expand more, unless someone else does it first!

 

[panta dokimazete]

model using 10 & expanding on the math:

8 - 5 = ?

9 - 5 = 4

4 + 8 = 12

12 - 10 = 2

2 + 1 = 3

 

[panta dokimazete]

Taking the 10 model and "formulizing" it:

8 - 5 = ((((9-5) + 8) - 10) + 1)

8 - 5 = (((4 + 8) - 10) + 1)

8 - 5 = ((12 - 10) + 1)

8 - 5 = (2+1)

8 - 5 = 3

 

[panta dokimazete]

I think this is how you could express the relationship:

Where x and y is < 10 and x > y

x - y = ((((9-y) + x) - 10) + 1)

just add zeros to the ten and have as many "9"s to increase the "power"...

example:

Where x and y is < 100 and x > y

x - y = ((((99-y) + x) - 100) + 1)

clear as mud?

BTW: somebody check my logic, please!

 

[panta dokimazete]

So here is the trick, I think...

Where x and y is < 10 and x > y

x - y = ((((9-y) + x) - 10) + 1)

The bolded part could also be formulated as:

x - y = ((((-y+9) + x) - 10) + 1)

so... +9+1 = 10 right? now see in italics

x - y = ((((-y+9) + x) - 10) + 1)

you basically are doing (9+1) - 10 or 10 - 10 = 0 - a net zero sum

 

[jfschultz]

Have any of you logical, brilliant minds ever heard of this? Someone in my church showed it to me tonight and asked if I could help her figure it out so that she can receive bonus points in her college course.

Create a subtraction problem (have a subtrahend with three digits and a minuend with three digits, but don't subtract it yet). Subtract the subtrahend from 999. Then add the difference to the original minuend. When you get the sum cross out the number in the thousands place and add the number that you cross out. That gives you the answer to the original subtraction problem.

Example problem: 727 - 397 (that's the original subtrahend and minuend)

999 - 397 = 602

602 + 727 = 1329 (cross out the 1)

329 + 1 = 330 (the answer to your original problem)



How or why does it work everytime? Thanks!

Ahh, this brings back memories from the 1970's and working with PDP-8 and PDP-12 computers, which had add, but not subtract instructions. One would take the subtrahend and do a complement and increment instruction. This would change all the 1's to 0's and all the 0's to 1's and then add one to form the 2's complement negative of the number. Then that is added to the minuend to get the result. With a fixed length of bits the last carry would fall off the end of the accumulator.

Your scheme would also work this way.

999 - 397 = 602 // this forms the 9's complement

602 + 1 = 603 // add 1 to get the 10's complement

603 + 727 = 1330 (cross out the 1)

answer 330