Question

Hello, Ive been trying to find the smallest number that is evenly divisible by numbers 1-20 with python. Reply if you have some knowledge on problems like this.

My attempt: x=2520
>>> y=1
>>> while y<20:
y+=1
x+=1
if x>2520 and x%y is 0:
print str(x)+ "is evenly divisible with"+str(y)
else:
print "This code was uneffective"

Answer 1

Here is the first somewhat naive option, which gets to 18 fairly quickly but takes a long time to get to 20. I suggest running it and looking at the output.

div_min = 0
num = 1
while True:
divisable = True
i = 1
while i < 21 and divisable:
if num % i != 0:
divisable = False
elif i > div_min:
div_min = i
print("First num divisable by 1 to", i, "is", num)
i += 1
if i == 21:
print("Divisable by 1 to 20:", num)
break
num += 1

After looking at the output you realize that each new divisable is a multiple of the previous one so instead of incrementing by 1 each time on last line, increment by previous divisable. Also, no sense checking all previous divisors so i = 1 becomes i = div_min +1. And since we can do much higher numbers now, let's implement max div so we can do more than just 1..20. Code below does 1..1000. Also comment out first print as we probably don't want to see all 1000.

div_min = 0
div_num = 1 #previous divisable
num = 1
max_div = 1000 #max divisible
while True:
divisable = True
i = div_min + 1 #only check divisors not previously checked
while i < max_div + 1 and divisable: #use max_div now
if num % i != 0:
divisable = False
elif i > div_min:
div_min = i
div_num = num #save previous divisable
#print("First num divisable by 1 to", i, "is", num)
i += 1
if i == max_div + 1: #use max_duv now
print("Divisable by 1 to", max_div, ":", num) #use max_div now
break
num += div_num #increment by previous divisable

Answer 2

Is it supposed to say "('Divisable by 1 to 20:', 1)" as an output?

Answer 3

The second one, if you have max_div = 20 will output "Divisable by 1 to 20 : 232792560"

Answer 4

Ok