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MIT 6.00 Intro Computer Science (OCW) 79 Online
OpenStudy (anonymous):

I am working on problem set 1 and have hit a bit of a wall. Can somebody help?

OpenStudy (anonymous):

What have you tried so far?

OpenStudy (anonymous):

primeCounter = 1 numbertoTest = 3 while primeCounter < 1000: for divisor in range (2, numbertoTest): if numbertoTest % divisor == 0: numbertoTest = numbertoTest + 2 else: divisor = divisor + 1 if numbertoTest != 0: primeCounter = primeCounter + 1 numbertoTest = numbertoTest + 2 print numbertoTest

OpenStudy (anonymous):

I think you should experiment a bit with how for loops work, the else: divisor = divisor + 1 section doesn't do anything. Also, why are you checking to see if numbertoTest doesn't equal zero? Where and why would your code set it to zero? Both those are tangential to the problem your having though. What's the algorithm you're using? Describe it in as much detail as possible.

OpenStudy (anonymous):

Looking at your first if/else loop. If the number is 9. Range will be 2,9. So...9/2 has a remainder. Div + 1. 9/3 = no remainder. so, add 2 to numerator. 11/3 = no remainder. Add 1 to divisor. 11/4 - no rem. 11/5 no remainder, etc, up to 11/10, then no remainder, so go to next IF statement where the counter is increased. Your algorithm is incorrect. You need a command to break the cycle, before you do all the calculations for the entire range of "divisor in range(2,numbertoTest).

OpenStudy (anonymous):

I made some typos...please forgive me - so text should be: Looking at your first if/else loop. If the number is 9. Range will be 2,9. So...9/2 has a remainder. Div + 1. 9/3 = no remainder. so, add 2 to numerator. 11/3 = remainder. Add 1 to divisor. 11/4 - rem. 11/5 remainder, etc, up to 11/10, then remainder, so go to next IF statement where the counter is increased. Your algorithm is incorrect. You need a command to break the cycle, before you do all the calculations for the entire range of "divisor in range(2,numbertoTest).

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