my code is off by 10 compared to other test cases, but not all the time, any ideas?
balance = 4157
annualInterestRate = 0.18
payment = 10
count = 0
while count < 12:
x = (x - payment) + (x*(annualInterestRate / 12))
count = count + 1
if count == 12:
while payforyear(balance) > 0:
payment = payment + 10
print ('Lowest payment: ' +str(payment) )
MIT 6.00 Intro Computer Science (OCW)
Stacey Warren - Expert brainly.com
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Why do you check if count == 12? Since the while loop will end anyway when the count reaches 12, you can just return the value!
if i recall correctly, if i didn't check count 12 and put the return x line in it returned the result after one iteration of the loop
Well, the main thing that is wrong is the formula (x-paymnet)+(x*(annualInterestRate/12))
If you check instructions it is (xipayment)*(1-(annualInterestRate/12))
This will fix your problem but stylistically using variables defined outside the function is bad practice unless absolutely necessary.
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Sorry just to be clear, which exercise are you trying to solve? I guess it's the second one!
That should read (x-payment)...
And (1+(... not minus
x = (x - payment) *(1+annualInterestRate / 12) should do the trick I guess
Yes, it does actually problem is I have no idea how, and why i got it wrong in the first place. I have looked at that area of the code more than a few times over the past 12 hours I have been working on it. I have this weird feeling that my solution is very very very far from other peoples solutions.
Yes and as chris said previously the if count==12 is not necessary as long as the return statement is lined up with the while, I tested it too.
yes thank you for that Chris, I am still getting my head around logic and python work flow. I was scratching my head with the while statement and indentation for a good couple of hours.
Indentation in python is very important to the meaning and this is a good reason for beginners to learn python because it forces good habits of indentation. Which isn't necessary in some languages but always recommended.