heather - 10 months ago 42

Python Question

I have the code

`num = 1`

num2 = 1

num3 = 1

list = []

list2 = []

list3 = []

def numCheck1 (num):

while num<1001:

if (num%3==0):

if (num%5==0):

print num

list.append(num)

num+=1

numCheck1(num)

break

else:

print "error 1"

else:

print "error 2"

num+=1

numCheck1(num)

total=sum(list)

print list

print total

def numCheck2 (num2):

while num2<1001:

if (num2%5==0):

print num2

list2.append(num2)

num2+=1

numCheck1(num2)

break

else:

print "error"

numCheck2(num2)

def numCheck3 (num3):

while num3<1001:

if (num3%3==0):

print num3

list3.append(num3)

num3+=1

numCheck1(num3)

break

else:

print "error"

numCheck3(num3)

total2 = sum(list2)

total3 = sum(list3)

overall = (total2 + total3) - total

print list2

print list3

print total2

print total3

print overall

As a basic summary of my code, I have 3 functions, and corresponding lists and variables for each of them. The first function checks for all multiples of 3 and 5 below and equal to 1000. The second checks for all multiples of 5 below and equal to 1000. The third checks for all multiples of 3 below and equal to 1000. The numbers that are multiples are added to the corresponding list, while the corresponding variable is incremented to allow the function to check all numbers. At the end, the program calculates 4 totals: the total of each of the lists, and a special total, which adds together the second two totals and subtracts the first to prevent overcounting. This is just the overarching structure of the program.

This program is supposed to solve this problem (not homework, just fun). The code is working (as far as I know; the first function definitely works) but it keeps crashing the compiler (I am using an online compiler, repl. I am wondering if there are any ways to make this code more efficient.

Thanks!

Answer

`numCheck1`

: Since *all* multiples of 3 and 5 are *also* multiples of 15, you can simply print out each item in `range(0, 1001, 15)`

(counts by 15)

`numCheck2`

: `range(0, 1001, 5)`

(counts by 5) should already be all multiples of 5 less than or equal to 1000.

`numCheck3`

: `range(0, 1001, 3)`

(counts by 3) is the same thing as above, simply with multiples of 3.

Source (Stackoverflow)