(12C) Modulo (the remainder after division )

[Gamo]

The modulo operation finds the remainder after division of one number by another.

Example: X ENTER Y R/S

6 MOD 4 = 2
7 MOD 4 = 3
1235 MOD 25 = 10
91596 MOD 123 = 84
456654 MOD 104 = 94

Special one: 98765 MOD 43210 = 12345

Program:

Code:


STO 0
  ÷
FRAC
RCL 0
  x

With this small program I do not see Round Off Error yet.

Gamo

[Dieter]

(01-09-2018 06:03 AM)Gamo Wrote:  The modulo operation finds the remainder after division of one number by another.
...

Code:

STO 0
  ÷
FRAC
RCL 0
  x

With this small program I do not see Round Off Error yet.

There is more than you see in FIX 2. ;-)

7 mod 3 => 0,99999999
20 mod 7 => 5,999999999
200 mod 9 => 1,999999980
1E+9 mod 6 => 4,2

Or take you own examples:

91596 mod 123 => 82,99999640
456654 mod 104 => 93,99998400
98765 mod 43210 => 12344,99999

You could add 0,4 + INTG to round the results to an integer which would help in most cases. But this limits the output to integers so that 123 mod 2,5 = 0,5 cannot be calculated. And in a way this also is like cheating. ;-)

As already noted, using b*frac(a/b) is not a good idea at all as it causes the shown roundoff errors. A better method is a – b*int(a/b). However, this requires some more steps if you want to preserve the stack:

Code:

01 X<>Y
02 STO 0
03 X<>Y
04 /
05 LASTX
06 X<>Y
07 INTG
08 x
09 RCL 0
10 X<>Y
11 -
12 GTO 00

This calculates all above examples correctly.

All these methods are limited by cases where due to limited precision a/b is returned as an integer although it actually is not.
Example: With 10-digit precision 8E+9 / 3 = 2666666667 so that the modulus is returned as 0 (frac method) or –1 (int method) instead of 2.

Dieter

[Gamo]

Thank You Dieter
Your program algorithm is very precise.

Gamo

[Gamo]

Another MOD program for 12C

Code:


ENTER
ENTER
-
Roll Down
X<>Y
LSTx
/
INTG
x
-

Example:
MOD (156511533, 1546223)

156511533 ENTER 1546223 R/S result 343010

Gamo

[Gamo]

This little MOD function routine must be the perfect one of all, I hope.

Procedure:

x mod y = remainder

X [ENTER] Y [R/S] display Answer

Program:

Code:


01    ÷
02  LSTx
03  X<>Y
04  ENTER
05  INTG
06    -
07    x
08  FIX 0

Example:

98765 mod 4321

98765 [ENTER] 4321 [R/S] display 3703

Gamo

[Joe Horn]

Although that routine is very short, it suffers from roundoff error. It even gets the answer wrong for your example (as can be seen if you set FIX 9, as Dieter pointed out previously in this thread). The 10-step routine you posted previously gets the correct answer.