- anonymous

been at problem set one for days now. i have a number of elements but cant get them to work together.

- jamiebookeater

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- anonymous

anyone wanna help?

- anonymous

what have you got so far?

- anonymous

haha well now that i look i feel like i have a lot less than i thought i did.
basically i have this:
for n in range(1000):
if n < 2: #number must be larger than two
pass
elif n%2==0: #number must not be even
pass
elif not n & 1:
pass
for x in range(3, int(n**0.5)+1, 2):
if n % x == 0:
pass
but i feel like i am doing many many many things wrong

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## More answers

- anonymous

for instance, should i be using a for loop?
and should i be using a range?
i get that i should get the program to count prime numbers 1000 times (or 999 not including 2) but i have no idea how to make that happen

- anonymous

well, let's try to describe the problem. What is the program going to do?

- anonymous

count 1000 prime numbers

- anonymous

ok. So if we can find prime numbers, it's easy to count them, right? that's just an int variable that gets one added every time we find a prime.

- anonymous

okay...

- anonymous

So we need to start at a number,
check to see if it's prime, and
If it is prime,
add one to the counter
otherwise
move to the next number

- anonymous

i see. so it will count to 1000 and show the prime number corresponding to the count?

- anonymous

right. the 1000th prime number

- anonymous

which is 7919, for when you need to check output

- anonymous

So say we have a number. We know we can get a number, we know we can increment a counter if it turns out to be prime, but for now we've got this number. What process can we use to tell if it's prime?

- anonymous

first make sure it's positive, then an integer, then odd, and then indivisible by... 2, 3, or 5?

- anonymous

how do we increment a counter if it turns out to be prime?

- anonymous

first, we assign a counter variable, as in:
ctr = 0
then, if the number is prime, we add to it:
if numberisprime:
ctr +=1
the += means 'equals itself plus,' so that's like saying ctr = ctr+1

- anonymous

okay. so then are my criteria for prime correct?

- anonymous

well, what if the number is divisible by 7? or 11? or a really high prime number that neither of us can think of?
One way to be absolutely sure is to just check. Try dividing the number by every number less than it.

- anonymous

We're using computers. We don't have to be clever, because we can do 2.9 million operations in a second.

- anonymous

so what function will divide by all integers below n?

- anonymous

well, a for loop would work. have you used those?

- anonymous

ive been playing around with them, went through this chapter last night - http://en.wikibooks.org/wiki/Python_Programming/Loops

- anonymous

So you know that with a for loop you can specify a range

- anonymous

what if you used a for loop with the range set to the number you wanted to test? That way it would iterate over every number less than the number you wanted to test.

- anonymous

but what number do i want to test? i could throw the 1000th prime in but wouldnt that defeat the purpose?

- anonymous

So if
numberToTest = 7
and your range is
for divisor in range (2,numberToTest)
then the for loop will start with x = 2, then x=3...up to 6.

- anonymous

oh, you want to start with a low number, and test every number going up until you collect 1000 primes

- anonymous

okay so i would start with a loop that lists primes and tell it to stop after it has counted 1000?

- anonymous

right:
while we haven't found as many primes as we need:
Check the next number
if the next number is prime
add one to the counter
When we have enough primes
Print the last prime we found.

- anonymous

while we haven't found as many primes as we need: # So we need a counter variable
Check the next number ##we need a number to test variable
if the next number is prime
add one to the counter
When we have enough primes ##we also need a goal variable
Print the last prime we found.

- anonymous

So we can start out saying
NumToTest = The number we're going to test for primeness
Divisor = The number we're going to divide NumToTest by to see if it's prime
Goal = the number of primes we want to find
ctr = the number of primes we've found

- anonymous

Now we can use those things as if they were actual numbers. So we can say
while ctr < Goal:
and that means "while we haven't found enough primes"

- anonymous

while ctr = goal: ##"while we haven't found enough primes
while Divisor < NumToTest ## while the number we're trying to divide by is smaller than the number we're testing

- anonymous

so ive got: NumToTest = 2
Divisor =
Goal = 1000
ctr = 0
while ctr < Goal:
but i do not know what to set divisor to

- anonymous

or should i start at 3?

- anonymous

while ctr = goal: ##"while we haven't found enough primes
while Divisor < NumToTest

- anonymous

We don't want to try to divide by 1--nothing would seem to be prime. We don't have to start at two if we only use odd numbers, and if we start out testing an odd number and always add two to it, we'll never have to test an even number. So the divisor should probably start at 3.

- anonymous

ahh thats nice.

- anonymous

I'm going to head home now. I'll be on later, but see what you can do to fill in that outline of a function. Make sure you start out with NumberToTest being odd, and make sure you increment things consistently--that is, whenever you reset the divisor, make sure you reset it to the same number, and whenever you add something to NumberToTest to get the next one, make sure you add the same number.

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