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 ▼ Thomas Klemm Senior Member Posts: 735 Threads: 34 Joined: May 2007 12-10-2009, 08:44 PM Recently I purchased a new mobile phone (SAMSUNG SGH-C180) and as always I was curious about the features of the calculator. The manual was short with the description: Quote: Perform basic arithmetic functions. Here's the menu: ```* . # + - x / % ( ) ``` Just an ordinary four-banger, I was thinking first. But then I realized it conforms to algebraic notation: ```2+3*4 = 14 (2+3)*4 = 20 ``` Up to 16 places can be displayed, however at most 6 places after the decimal point: ```1000*1000*1000*1000*1000 = Out of range 1000*1000*1000*1000*1000-1 = 999999999999999 (1000*1000*1000*1000*1000*1000-1000)/1000 = 999999999999999 (1000*1000*1000*1000*1000-1)/1000 = 999999999999.999 355/113 = 3.141593 355000000000/113 = 3141592920.353982 1234567890+0.123456 = 1234567890.123456 12345678901+0.123456 = 12345678901.1235 123456789012+0.123456 = 123456789012.123 9876543210+0.123456 = 9876543210.123455 98765432109+0.123456 = 98765432109.1235 987654321098+0.123456 = 987654321098.123 ``` Extracting the square root can be done iteratively since you can edit the last computation: ```(3/1+1)/2 = 2 (3/2+2)/2 = 1.75 (3/1.75+1.75)/2 = 1.732143 (3/1.732143+1.732143)/2 = 1.732051 (3/1.732051+1.732051)/2 = 1.732051 ``` Error messages are nice: ```1/0 = Can't divide by zero ``` But then I stumbled over the following: ```23+17% = Syntax Error ``` Only after a while I found out how to use the percentage operation: ```23%17 = 6 355%113 = 16 ``` Who'd consider modulo a basic arithmetic function? Have you ever noticed the use of % for this operation in a calculator? (I know it's common in many computer languages.) Kind regards Thomas Edited: 10 Dec 2009, 8:54 p.m. ▼ hugh steers Senior Member Posts: 536 Threads: 56 Joined: Jul 2005 12-10-2009, 09:03 PM im thinking the calculator in your phone is working in binary IEEE754 and compiled with the usual runtime mathlib. you could check by calculating, say 1/17 and then multiplying and subtracting the significant figures. if it's binary, the numbers will go on and on and you will get garbage after 16 digits. if it's decimal, you will hit the zeros. ▼ Thomas Klemm Senior Member Posts: 735 Threads: 34 Joined: May 2007 12-15-2009, 04:35 PM First I was in doubt since e.g. 1/17*17-1 would just result in 0 as expected. Only after a while I came up with the following expression: ```(1-(1000000-999999.999999)*1000000)*10000000000000 ``` Using Python results in: ```-76144933.700561523 ``` While the calculator of my phone would yield: ```-76144933.700562 ``` I couldn't get any more places since the input field is limited to 50 characters. So it seems intermediate results are calculated using binary while the result is rounded to 6 places after the decimal point. Of course HP-calculators using BCD don't encounter problems with this expression.

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