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

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- Dec 21st 2009, 10:12 AMhebbymathematical operations
how it is possible to remove the last three-digit of a number N using two mathematical operations

- Dec 21st 2009, 10:28 AMPlato
- Dec 21st 2009, 10:34 AMtonio
- Dec 21st 2009, 10:59 AMhebby
well that well only get u a decimal with all the digits still there

- Dec 21st 2009, 11:00 AMhebby
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, 11:01 AMhebby
N= whole number...integer

- Dec 21st 2009, 11:02 AMhebby
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, 02:47 PMtonio

No operations needed: if the number is $\displaystyle N=a_1a_2\ldots a_n\,,\,\,a_i\in\{0,1,2,\ldots,9\}$ , then take $\displaystyle N'=a_1\ldots a_{n-3}$ (Giggle)

Or more seriously: take $\displaystyle \left[\frac{N}{1000}\right]$, as already proposed by Plato.

Tonio - Dec 21st 2009, 02:52 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, 03:01 PMPlato
- Dec 21st 2009, 03:03 PMhebby
no i never saw that before, but i can see how that would work now

- Jan 5th 2010, 03:37 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, 09:07 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, 09:10 PMhebby
Could you show a numerical example on how that might work?

- Jan 5th 2010, 09:17 PMDrexel28