how it is possible to remove the last three-digit of a number N using two mathematical operations

Printable View

- Dec 21st 2009, 11:12 AMhebbymathematical operations
how it is possible to remove the last three-digit of a number N using two mathematical operations

- Dec 21st 2009, 11:28 AMPlato
- Dec 21st 2009, 11:34 AMtonio
- Dec 21st 2009, 11:59 AMhebby
well that well only get u a decimal with all the digits still there

- Dec 21st 2009, 12:00 PMhebby
ya basically from the right....ie the last 3 digits....the operations u mentioned are all valid.....but a general method...to remove the last 3 digits from any number N

- Dec 21st 2009, 12:01 PMhebby
N= whole number...integer

- Dec 21st 2009, 12:02 PMhebby
eg) 12345=N and u end up getting 12....but minus last 3 divide by 100 is not valid as this can be applied once we know the numbers....but we actually dont know the number N

- Dec 21st 2009, 03:47 PMtonio
- Dec 21st 2009, 03:52 PMhebby
hey thanks, but how would the N divided by 1000 work? as you will end up with a number with decimals, how would you remove those?

- Dec 21st 2009, 04:01 PMPlato
- Dec 21st 2009, 04:03 PMhebby
no i never saw that before, but i can see how that would work now

- Jan 5th 2010, 04:37 PMhebby
Hi

I was think in the lines of

1) Multiply by 10^ -3

2) The we end up with a decimal, how can I remove the decimal from the integer?

Any ideas? - Jan 5th 2010, 10:07 PMBingk
I was thinking ... would the floor function be considered as an operation? Just curious about that

Could you also try ( N - ( N (mod 1000) ) ) / 1000? Basically, N minus the remainder when you divide by 1000, then divide that by 1000 .... - Jan 5th 2010, 10:10 PMhebby
Could you show a numerical example on how that might work?

- Jan 5th 2010, 10:17 PMDrexel28