
Anyway To Calculate Remainder?
#1
Posted 16 April 2006 - 06:31 AM
Finding out remainders is a standard thing to do but is there a function to find out remainders?
I dont' think the simplied fraction to the lowest term is useful for big and high powered numbers.
Is there a way to do it on the fx-991MS S-V.P.A.M. calculator?
the link to the manual is here: http://ftp.casio.co....5MS_991MS_E.pdf
Thanks
#2
Guest_Guest_*
Posted 19 April 2006 - 10:09 AM
I'm taking a module on mod and congruency, primitive roots and index arithmetic.
Finding out remainders is a standard thing to do but is there a function to find out remainders?
I dont' think the simplied fraction to the lowest term is useful for big and high powered numbers.
Is there a way to do it on the fx-991MS S-V.P.A.M. calculator?
the link to the manual is here: http://ftp.casio.co....5MS_991MS_E.pdf
Thanks
Yes, it is very easy ...
OriginalValue - Int(OriginalValue/Divisor)*Divisor
#3
Posted 19 April 2006 - 11:04 AM
Yes, it is very easy ...
OriginalValue - Int(OriginalValue/Divisor)*Divisor
pardon my ignorance but which buttons do I actualy have to press to create that function ?
#4
Guest_Guest_*
Posted 19 April 2006 - 11:59 AM
pardon my ignorance but which buttons do I actualy have to press to create that function ?
duh - The ones on the front

#5
Guest_samuel_*
Posted 19 April 2006 - 02:15 PM
First you calculate Original/divisor; See what is the integer part
Then use the integer part to multiplies with the divisor. Finally Use the original to subtract the products.
For example, say to find remainder when 73 is divided by 8
So you find 73/8 = 9.125, and the integer part is 9.
Now find 9*8=72
So your remainder is 73-72=1
OK?

#6
Guest_Fatu_*
Posted 02 July 2006 - 10:11 PM
#7
Posted 03 July 2006 - 05:55 PM
#8
Posted 04 July 2006 - 08:59 AM
#9
Posted 04 July 2006 - 09:31 AM
To find the remainder on 991MS we should use the algorithm. 991MS is not programmable and has no INT or FRAC functions.
#10
Posted 04 July 2006 - 06:10 PM

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original
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Should make it. The idea is to use Rnd and Fix 0 to get the integer part.
The
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Nice challenge

#11
Posted 01 August 2006 - 12:33 PM
I'm taking a module on mod and congruency, primitive roots and index arithmetic.
Finding out remainders is a standard thing to do but is there a function to find out remainders?
I dont' think the simplied fraction to the lowest term is useful for big and high powered numbers.
Is there a way to do it on the fx-991MS S-V.P.A.M. calculator?
the link to the manual is here: http://ftp.casio.co....5MS_991MS_E.pdf
Thanks
May be a little late but how about this method?
Press MODE MODE 3 to enter BASE mode. Press DEC if necessary. The calculator should show d.
Now suppose you want to find the remainder when 24 is divided by 5. You would press 24 - 24 ? 5 X 5 =
The calculator should show the remainder 4.
In BASE mode the calculator automatically performs integer division. This is also true on the FX-3650P/3950P and you can use this property in writing programs.
Press MODE 1 to return to COMP mode.
#12
Guest_blabla_*
Posted 20 November 2007 - 04:16 PM
I can't use anyone cos my number is too big like 88^7 mod 187.....
So how to do?
#13
Guest_ujazz_*
Posted 10 March 2008 - 04:56 AM
http://web.ew.usna.e...ook/node27.html
#14
Guest_vh_*
Posted 01 October 2008 - 08:47 PM
#15
Guest_Naman Srivastava_*
Posted 30 December 2008 - 05:33 PM
I'm taking a module on mod and congruency, primitive roots and index arithmetic.
Finding out remainders is a standard thing to do but is there a function to find out remainders?
I dont' think the simplied fraction to the lowest term is useful for big and high powered numbers.
Is there a way to do it on the fx-991MS S-V.P.A.M. calculator?
the link to the manual is here: http://ftp.casio.co....5MS_991MS_E.pdf
Thanks
#16
Guest_Naman Srivastava_*
Posted 30 December 2008 - 05:39 PM
Actually i was telling the solution to fing mod of any number on fx991MS Calculator.
its simple
Change the display to a b/c
and divide the no using the fraction symbol.
For Exampel
30/13 = 2 4/13
and there u go.... 4 is ur remainder !!!
#17
Guest_Nicholas_*
Posted 20 November 2009 - 03:54 AM
#18
Guest_Rino_*
Posted 24 May 2010 - 11:48 AM
Using the algorithm Original-Int(Original/divisor)*divisor
First you calculate Original/divisor; See what is the integer part
Then use the integer part to multiplies with the divisor. Finally Use the original to subtract the products.
For example, say to find remainder when 73 is divided by 8
So you find 73/8 = 9.125, and the integer part is 9.
Now find 9*8=72
So your remainder is 73-72=1
OK?
#19
Guest_Rino_*
Posted 24 May 2010 - 11:49 AM
#20
Guest_b.kaarthick_*
Posted 25 October 2010 - 05:45 AM
its wrongYes, it is very easy ...
OriginalValue - Int(OriginalValue/Divisor)*Divisor
#21
Guest_Guest_*
Posted 25 October 2010 - 05:47 AM
#22
Guest_Sagar_*
Posted 08 December 2010 - 01:16 PM
1, 3 ,4 which is 4*1=4 and 7-4=3 remainder...
#23
Guest_Felix_*
Posted 01 February 2011 - 04:25 PM
if you want to round with 2 digits after the comma
int (number * 100) / 100
with 3 digits after comma:
int (number * 1000) / 1000 and so on
#24
Guest_Mahesh_*
Posted 23 May 2011 - 02:48 PM
but ders a "a b/c" button near top left corner..
after finding say 7/3.. press that "a b/c" , so the answer will be the "b" value.. (here ans =4) that is the middle value..
8/4 gives 2, that is no decimal value, then mod value is 0..
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