Using MOD

[Robert E. Carneal]

If I use MOD in Powerbasic (Dos) for the number of

1960092520030605, I get 115.

If I do the same thing in Lotus or Excel, I get 31 instead.

Yes, I know R = Num1 MOD Num2 yields the remainder, but shouldn't
the remainder be same no matter which application is doing the
calculating? I am concerned I do not understand MOD as well
as I had believed I did. I wonder if my number is too large for
accuracy?

Thanks.

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[Mel Bishop]

What are you MOD'ing it by?


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[Robert E. Carneal]

253

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[George Bleck]

What data type is your variable defined as?

I tried using MOD(1960092520030605, 253) in ExcelXP and I get an error.

I then tried the formula for MOD as such...1960092520030605 - (253 * INT(1960092520030605 / 253))
In ExcelXP (in a single cell) and got 21. I also tried TRUNC instead of INT and got the same answer.

I then tried the same formula step by step in a different cells.

1960092520030605 / 253 = 7747401264943.08
253 * 7747401264943 = 1960092520030580 (!)
1960092520030605 - 1960092520030580 = 20

Using the same function above with Windows Calculator.

1960092520030605 / 253 = 7747401264943.102766798418972332
253 * 7747401264943 = 1960092520030579
1960092520030605 - 1960092520030579 = 26

Using the below PB program (granted I tested under PBCC 3.02) I get the following...

FUNCTION PBMAIN
	LOCAL a1 AS BYTE
	LOCAL a2 AS INTEGER
	LOCAL a3 AS WORD
	LOCAL a4 AS LONG
	LOCAL a5 AS DWORD
	LOCAL a6 AS SINGLE
	LOCAL a7 AS QUAD
	LOCAL a8 AS DOUBLE
	LOCAL lngModVal AS LONG

	lngModVal = 253

	a1=1960092520030605
	a2=1960092520030605
	a3=1960092520030605
	a4=1960092520030605
	a5=1960092520030605
	a6=1960092520030605
	a7=1960092520030605
	a8=1960092520030605

	PRINT a1 MOD lngModVal, a1 - (lngModVal * FIX(a1/lngModVal)), a1 - (lngModVal * INT(a1/lngModVal))
	PRINT a2 MOD lngModVal, a2 - (lngModVal * FIX(a2/lngModVal)), a2 - (lngModVal * INT(a2/lngModVal))
	PRINT a3 MOD lngModVal, a3 - (lngModVal * FIX(a3/lngModVal)), a3 - (lngModVal * INT(a3/lngModVal))
	PRINT a4 MOD lngModVal, a4 - (lngModVal * FIX(a4/lngModVal)), a4 - (lngModVal * INT(a4/lngModVal))
	PRINT a5 MOD lngModVal, a5 - (lngModVal * FIX(a5/lngModVal)), a5 - (lngModVal * INT(a5/lngModVal))
	PRINT a6 MOD lngModVal, a6 - (lngModVal * FIX(a6/lngModVal)), a6 - (lngModVal * INT(a6/lngModVal))
	PRINT a7 MOD lngModVal, a7 - (lngModVal * FIX(a7/lngModVal)), a7 - (lngModVal * INT(a7/lngModVal))
	PRINT a8 MOD lngModVal, a8 - (lngModVal * FIX(a8/lngModVal)), a8 - (lngModVal * INT(a8/lngModVal))

	WAITKEY$
END FUNCTION

a1 =  141           141           141
a2 =  240           240           240
a3 =  240           240           240
a4 = -167          -167           86
a5 =  220           220           220
a6 =  115           115           115
a7 =  26            26            26
a8 =  26            26            26

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Every day I try to learn one thing new,
but new things to learn are increasing exponentially.
At this rate I’m becoming an idiot faster and faster !!!
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George W. Bleck
Lead Computer Systems Engineer
KeySpan Corporation
My Email



[This message has been edited by George Bleck (edited June 05, 2003).]

[Mel Bishop]

In PB/DOS I used:

a&& = that number
print;a&& mod 253 and got the result of 26.

Just for larks, I then:
print;a&& / 253 and received a remainder result of .10278

Go figgure.

Now, admittedly, we are dealing with a really big number and I'm
sure there is an explaniation for this....
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[This message has been edited by Mel Bishop (edited June 05, 2003).]

[Michael Mattias]

Just for larks, I then:
print;a&& / 253 and received a remainder result of .10278

Go figgure.

OK, Figgured.

"/" does not create a 'remainder result.'

Neither does "\" for that matter.

Only MOD creates a 'remainder.'

MCM

[Paul Dixon]

Dear all,
the answer is 26.

The number 1960092520030605 is too big to be correctly represented except as QUAD or EXT types so make sure you specify the right type.

The line:
PRINT 1960092520030605 MOD 253

correctly prints 26 as the result in PBDOS3.5, PBCC1.0, UBASIC, and the numbers also work on Windows calculator in Win98se.

Robert,
you get 115 because you have assigned the number to a real (a!=..) which only holds about 7 digits.

Mel,
<<Go figgure.>>

Go figure what? That's what you should get. 0.10278 * 253 = 26 as we'd expect.

Paul.

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[This message has been edited by Paul Dixon (edited June 06, 2003).]