It's been a while since I last wrote some lolcode, so I felt like tackling Project Euler #1.

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.

This is the code/script I wrote to solve this:

HAI 1.2
    VISIBLE "HAI, PROJEK LOLLER ONE!!"
    
    I HAS A LIMIT ITZ 1000
    I HAS A TOTAL ITZ 0
    I HAS A CHEEZ ITZ 3
    I HAS A BURGER ITZ 5
    I HAS A CHEEZBURGER ITZ PRODUKT OF CHEEZ AN BURGER
    
    HOW IZ I ADDTOTAL YR VALUE
        I HAS A RESULT ITZ SUM OF VALUE AN TOTAL
        FOUND YR RESULT
    IF U SAY SO
    
    
    IM IN YR MIND UPPIN YR NUMBER TIL BOTH SAEM NUMBER AN LIMIT

        I HAS A PICKLE ITZ FAIL
        
        BOTH SAEM 0 AN MOD OF NUMBER AN CHEEZBURGER
        O RLY?, YA RLY
            PICKLE R WIN
        NO WAI
            BOTH SAEM 0 AN MOD OF NUMBER AN CHEEZ
            O RLY?, YA RLY
                PICKLE R WIN
            NO WAI
                BOTH SAEM 0 AN MOD OF NUMBER AN BURGER
                O RLY?, YA RLY
                    PICKLE R WIN
                OIC
            OIC
        OIC
        
        BOTH SAEM PICKLE AN WIN
        O RLY?, YA RLY
            TOTAL R I IZ ADDTOTAL YR NUMBER MKAY
        OIC
        
    IM OUTTA YR MIND
    
    VISIBLE SMOOSH "TEH ANSWER IZ " AN TOTAL MKAY
    VISIBLE "DOWN WITH PROJEK LOLLER ONE!"

KTHXBYE

This code can be executed on CodingGround, and produces this output:

HAI, PROJEK LOLLER ONE!!

(spoiler)

TEH ANSWER IZ 233168

DOWN WITH PROJEK LOLLER!

Is the fizzbuzz-like logic appropriate, or I fell in the "easy" trap? Would there be a way to collapse all the conditions into one and make a shorter script? Or is this algorithm a bad/inefficient one?