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Ti89 Titanium too few arguments

by Spagnum
Tags: ti89, titanium, too few arguments
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Spagnum
#1
Oct29-11, 10:22 AM
P: 4
sorry for my noobieness, but I'm having problem with my mod( function on my Ti89 Titanium

On other calcs: 54 mod 10 = 4
like here: http://www.google.com/search?q=54+mod+10

But when i use the mod( function I either get "to few arguments" og just the result 54.mod10

can anyone point me to what I'm doing wrong?
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#2
Oct29-11, 12:08 PM
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Welcome to PF, Spagnum!

Have you tried mod(54,10)?
Spagnum
#3
Oct29-11, 12:15 PM
P: 4
Thank you!

yes I have - that just results in "Too few arguments"

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#4
Oct29-11, 12:34 PM
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Ti89 Titanium too few arguments

Quote Quote by Spagnum View Post
Thank you!

yes I have - that just results in "Too few arguments"
Ah well, I do not own a Ti89, but this page: http://www.prenhall.com/esm/app/grap...nu/number.html
shows that mod(-5,8) --> 3 should work, complete with screenshot.
Spagnum
#5
Oct29-11, 03:21 PM
P: 4
Yup I been to that page myself but I get the same result.(Too few arguments) I even reset the calculator, in case it had something to do with some settings.
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#6
Oct29-11, 03:26 PM
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Quote Quote by Spagnum View Post
Yup I been to that page myself but I get the same result.(Too few arguments) I even reset the calculator, in case it had something to do with some settings.
The one reason I can think it might say "too few arguments", is if there's a problem with the decimal point, in which case you'd have tried to calculate mod(54.10, ???)
You sure you used a comma?
Or perhaps you could try a semicolon?
Spagnum
#7
Oct29-11, 03:41 PM
P: 4
ofc... the comma.
I'm so use to working in Excel where . is the comma, and haven't used my calculator in ages.

Anyways it works like a charm now, thanks a lot for your help Serena!
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#8
Oct29-11, 03:43 PM
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You're welcome!
Borek
#9
Oct29-11, 03:46 PM
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On my TI-89 (but not Titanium) mod(54,10) gives correct 4.
jtbell
#10
Oct29-11, 04:23 PM
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The old UK versus USA, comma versus period switcheroo, eh?
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#11
Oct29-11, 04:32 PM
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Uhh I've never heard of using a period as an argument separator?
titaniu
#12
Feb18-12, 11:01 AM
P: 1
Hey I am also a noob and I need help.
I made this program to calculate the area of a sector and I keep getting "Too Few Arguments"
The is a theta and the 'arrow' is the arrow made by STO


area_sec()
Prgm
Local a,r,g,s,,k
Dialog
Request "Press 1 if you have rs and 2 if you have r"a
EndDlog
If a=1 Then
Dialog
Request "Enter r",r
Request "Enter s",s
EndDlog
expr(-r) 'arrow' r
expr(s) 'arrow' s
(1/2)*r*s 'arrow' k
Disp "Area is"&string(k)
Else If a=2 Then
Dialog
Request "Enter r",r
Request "Enter ",
Dialog
expr(-r) 'arrow' r
expr(s) 'arrow' s
(1/2)*r*r* 'arrow' k
Disp "Area is"&string(k)
Else
EndIf
DelVar a,r,s,,k
EndPrgm
SonikkuUSA
#13
Aug4-12, 04:39 PM
P: 1
Look at line 5 again:

Request "Press 1 if you have rs and 2 if you have r"a
You forgot a comma in between your "request" and your variable, like so:

Request "Press 1 if you have rs and 2 if you have r",a
Try that and see what you get.
nathanthegreat
#14
Mar7-14, 11:27 PM
P: 15
SonikkuUSA is right about the comma. Also, after the first "EndDlog", you have to add "expr(a)->a". This is because request (and any dialog function) is treated as a string, and we want a value. Also, in the first request, type"Request"Words",a,0. The zero will turn alpha-lock off in that box so you don't have to hit the alpha button every time you want to type 1.

Instead of using delvar at the end, you could just use local and define all of the variables at the beginning, too.


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