I am new to python. I am trying to write a program that counts a range of values but for multiples of 3 and 4 adds 3 and 4 to them resepectively and when the number is multiple of both adds that number to it. So for instance 1,2,3,4,5,6,7,8,9,10,11,12 should read in the end program as: 1,2,6,8,5,9,7,12,10,11,24
But my code is getting stuck somewhere.
for i in range (1,20):
if i%3==0 and i%4==0:
i=i+12
if i%3==0:
i=i+3
if i%4==0:
i=i+4
print i
This line has a typo
if i%3==0 and 1%4==0: # your second condition is 1 % 4 == 0, which is always false
I think you meant
if i%3==0 and i%4==0:
It is better (accurate and cleaner) if you use a different variable.
for i in range (1,20):
n = i
if i%3==0 and i%4==0:
n=i+12
if i%3==0:
n=i+3
if i%4==0:
n=i+4
print n
Now you will notice this fixed it for the 9 case, but not the 12 case! We now need to add the use of elif. Also, if a number is a multiple of 3 and 4, then it is also a multiple of their lowest common multiple, in this case 12. So you could re-write your first step to just check for multiples of 12. This gives us:
for i in range (1,20):
n = i
if i%12==0
n=i+12 # or possibly i + i
elif i%3==0:
n=i+3
elif i%4==0:
n=i+4
print n
The reason this works is because without the elif the i was getting added to multiple times. For example, with 9, you get to 9%3 ==0, True. Now i is set to 12. Next statement? 12%4 ==0 True. So another 4 is added.
Alternatively, if you would like to do some crazy python stuff:
for i in range(1, 20):
print i + next((n for n in (12, 4, 3) if i % n == 0), 0)
(Shout out to Jon Clements for this 2-liner answer)
You're allowing multiples cases (if conditions) for each iteration. I think you probably wanted to do exclusive cases.
Maybe something like:
for i in range(1, 20):
print i, "=>",
if i % 3 == 0 and i % 4 == 0:
i += i
elif i % 3 == 0:
i += 3
elif i % 4 == 0:
i += 4
print i
+=syntax as well. Instead ofi=i+4, you could usei+=4. It'll make it cleaner.