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Thread: Two Consoles

  1. #1

    Euler Problem No. 1

    The first problem for the Euler Project is as follows:

    If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5. 6. and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.

    So I wrote the code below, which gives 36,560. The Euler Project says this answer is incorrect. Can someone give me a hint about what's wrong? I honestly can't see it.

    USES "Console"           
    DIM i AS Integer         
    DIM Total AS Word VALUE 0
                             
    FOR i = 1 TO 999         
                             
    IF MOD(i, 3) = 0 THEN    
    Total += i               
    ELSEIF MOD(i, 5) = 0 THEN
    Total += i               
    END IF                   
                             
    NEXT                     
                             
    CONSOLE_WRITELINE Total  
    WAITKEY
    
    Last edited by Benjamin; 18-12-2023 at 13:05. Reason: Changed subject

  2. #2

    I changed the post.

    I changed the question in the post. Please see above.
    Last edited by Benjamin; 18-12-2023 at 13:41.

  3. #3

    Modified; same result

    I simplified the logic, but still got 36,560, which the Euler Project says is incorrect.

    USES "Console"                        
    DIM i AS Integer                      
    DIM Total AS Word VALUE 0             
                                          
    FOR i = 1 TO 999                      
                                          
    IF MOD(i, 3) = 0 OR MOD(i, 5) = 0 THEN
    Total += i                            
    END IF                                
                                          
    NEXT                                  
                                          
    CONSOLE_WRITELINE Total               
    WAITKEY
    
    What gives?

  4. #4
    Member
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    wrong type of variable

    Hi Benjamin

    The variable total is declared word

    Word is 16 bits numbers from 0 to 65535

    when i reaches 528 total is 65103 and for i = 530 total wraps to 97 instead of 65633

    Use
    dim total as dword
    
    , 32 bits wide , up to 4294967295


    Try this version:
    Uses "Console"
    DIM i AS Integer                     
    DIM Total AS word VALUE 0             
    DIM total_dw AS dword VALUE 0             
                                           
    FOR i = 1 TO 999                   
                                           
    IF MOD(i, 3) = 0 OR MOD(i, 5) = 0 THEN
    
    Total += i                            
    total_dw += i                            
    printl i, total, total_dw
     END IF                               
             if  i = 530  then WaitKey           'pause                     
    NEXT                                 
                                           
    CONSOLE_WRITELINE Total      , total_dw            
    PrintL "Press a key to end program"
    
    '---Wait for a key press
    WaitKey
    
    Regards

    DC

  5. #5
    Hi DC,

    Thanks so much for the detailed explanation. I learned a trick with those two variables. I'll have to remember that. Believe it or not, the first time I wrote it, I used Integer, but when I got a negative number, I knew I had wrapped around.

    I'm relieved to know that I'm not losing my mind

    Benjamin

  6. #6
    thinBasic author ErosOlmi's Avatar
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    10
    To store numbers it is better to always use DWORD, LONG or QUAD for integer numbers and DOUBLE or EXT (EXTENDED) for floating point numbers.
    Leave all other numeric data types for compatibility with data structure you need to use or external functions (DLLs) that specifically require them.

    Even when you just need a simple counter for a loop, better to use LONG/DWORD.
    They are much faster because native 32bit
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