# find the first number that is divided by 1 to 9 with 0 remainder
for i in range (10000):
x = i % 3
y = i % 4
z = i % 7
a = i % 5
b = i % 6
c = i % 8
d = i % 9
if x==y==z==a==b==c==d==0:
print (i)
As it mention, i wanted to find numbers which are divided by 1 to 9 numbers with 0 remainders.
I tried to make it dynamic like instead of 1 to 9 what if i search numbers which are divided by 1 to 20 numbers.
I have already repeated lots of code line for 1 to 9 numbers and i don't want to do that for more numbers.
I tried to make a list of 1 to 20 numbers and with loop tried to iterate but failed.
Thanks in advance.
Maybe run two loops: one for the number to check, one for the division test.
max_check = 10000
dividends = [3, 4, 7, 5, 6, 8, 9]
for n in range(1,max_check):
for dividend in dividends:
if n % dividend:
break
else:
print(f"{n} is evenly divided by {dividends}")
break
else:
print(f"No numbers found evenly divisible by {dividends} that are less than {max_check}")
There is no reason to test every value. If the result must be divisible for all numbers 1 through 9, the number will be divisible by 9. There is no reason to test 17 or 33 or 121.
def find_number(first, last):
"""Find first number that is divisible by all numbers in range first..last"""
num = last
while True:
for divisor in range(first, last):
if num % divisor != 0: # Failed test. Try next num
num += last
break # break out of for loop
else: # Gets called if for loop reaches end
print(num)
break; # Break out of while loop
find_number(5, 13)
Great! Thank you so much.
The second nested 'for' loop with 'else' was the key that i missed.
And of course learned that i should use 0 as a boolean even i was working with arithmetic. It optimized code so well.
Thank you
you can use
any()
for number in range (1, 10000):
if not any(number % divisor for divisor in range(3, 10)):
print(number)
break # break out of loop early
else:
print('no such number found')note that probably you need to find more elegant (from math perspective) solution, not trying to brute-force it
Wow!
TIL any() function.
Thank you so much.