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Anyway To Calculate Remainder?

Started by Mankul , Apr 16 2006 06:31 AM

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#1 Mankul

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Posted 16 April 2006 - 06:31 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

#2 Guest_Guest_*

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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 Mankul

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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_*

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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 :huh:

#5 Guest_samuel_*

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Posted 19 April 2006 - 02:15 PM

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? :)

#6 Guest_Fatu_*

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Posted 02 July 2006 - 10:11 PM

Sorry, but i couldn't find the int button in the front pad. I understand that the question is if there exists an "int" function or "mod" function in this calculator model..., or at least a simple way to emulate it... ?

#7 Andy.Davies

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Posted 03 July 2006 - 05:55 PM

it isnt in the front pad, INT is in OPTN > NUM on the calc

#8 Scratty

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Posted 04 July 2006 - 08:59 AM

So, it doesn't have a Frac command? Otherwise it's just den*Frac(num/den) if memory serves.

#9 samuel

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Posted 04 July 2006 - 09:31 AM

I think you have mis interpret the original question...it was about finding remainder on the scientific calc 991MS or 991ES, not CFX or FX series. Those options you have suggested would not work.

To find the remainder on 991MS we should use the algorithm. 991MS is not programmable and has no INT or FRAC functions.

#10 TyYann

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Posted 04 July 2006 - 06:10 PM

[Mode] <span class=1' /> (Fix) [0]
original [/] divisor [=]
<span class=Shift' /> [Rnd]
[-] original [/] divisor [=]
<span class=×' /> divisor [=]
<span class=Shift' /> [Rnd]
[0] [-] <span class=Shift' /> [Ans] [=]

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

The [0] [-] <span class=Shift' /> [Ans] [=] part is to get the sign right.

Nice challenge ;)

#11 James Bond 007

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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_*

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Posted 20 November 2007 - 04:16 PM

hey.............

I can't use anyone cos my number is too big like 88^7 mod 187.....
So how to do?

#13 Guest_ujazz_*

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Posted 10 March 2008 - 04:56 AM

Well, you could use the repeated squaring algorithm to simplify the calculation a bit

http://web.ew.usna.e...ook/node27.html

#14 Guest_vh_*

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Posted 01 October 2008 - 08:47 PM

The Rnd function doesn't work, thats all, folks.

#15 Guest_Naman Srivastava_*

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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_*

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Posted 30 December 2008 - 05:39 PM

Sorry folks i posted the query agn by mistake..

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_*

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Posted 20 November 2009 - 03:54 AM

Good try but calculators always simplify fractions, so if the fraction part is 4/6 it will show 2/3 instead which will make this method have as many steps to compute as some of the other methods mentioned

#18 Guest_Rino_*

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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_*

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Posted 24 May 2010 - 11:49 AM

thanx a lot 4 this ....!!!!!

#20 Guest_b.kaarthick_*

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Posted 25 October 2010 - 05:45 AM

Yes, it is very easy ...

OriginalValue - Int(OriginalValue/Divisor)*Divisor

its wrong

#21 Guest_Guest_*

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Posted 25 October 2010 - 05:47 AM


#22 Guest_Sagar_*

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Posted 08 December 2010 - 01:16 PM

there is one button ab/c beneath CALC... if u want to calculate 7 mod 4 then... u should press 7 then ab/c 4 and equal to
1, 3 ,4 which is 4*1=4 and 7-4=3 remainder...

#23 Guest_Felix_*

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Posted 01 February 2011 - 04:25 PM

Do it this way:

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_*

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Posted 23 May 2011 - 02:48 PM

there is no separate key as such..
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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