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Start your free trialVictor Katsande
3,094 PointsMADE A COUPLE OF CORRECTIONS. STILL TELLING ME TASK ONE IS NO LONGER PASSING.
HERE IS THE ORIGINAL QUESTION
Alright, last step but it's a big one.
Make a while loop that runs until start is falsey.
Inside the loop, use random.randint(1, 99) to get a random number between 1 and 99.
If that random number is even (use even_odd to find out), print "{} is even", putting the random number in the hole. Otherwise, print "{} is odd", again using the random number.
Finally, decrement start by 1.
I know it's a lot, but I know you can do it!
import random
def even_odd(num):
# If % 2 is 0, the number is even.
# Since 0 is falsey, we have to invert it with not.
return not num % 2
start = 5
while start > 0:
num = random.randint(1, 99)
if even_odd(num) == 0:
print("{} is even".format(num))
else:
print("{} is odd".format(num))
start -+ = 1
1 Answer
Robert Ionut Muraru
Full Stack JavaScript Techdegree Graduate 22,194 PointsThe below syntax is not correct:
start -+ = 1
You want to use:
start -= 1
You need no spaces between minus and equal and the plus sign should not be there. You will then have to check the logic in the if / else statements, as it is printing the opposite of what is required:
if even_odd(num) == 0:
print("{} is even".format(num))
- the above is the same as saying to python: if even_odd() is False...print out that the number is even. Remember that 0 is falsey. The even_odd() function checks if the number is even and returns True if it is even.
Malvin Mahati
7,555 PointsMalvin Mahati
7,555 Pointsthis should work
def even_odd(num): # If % 2 is 0, the number is even. # Since 0 is falsey, we have to invert it with not. return not num % 2
def start(): start = 5
while start: num = random.randint(1, 99) if even_odd(num): print('{} is even'.format(num)) else: print('{} is odd'.format(num)) start -=1