finding multiples of a number in Python

Does this do what you want?

print range(0, (m+1)*n, n)[1:]

For m=5, n=20

[20, 40, 60, 80, 100]

Or better yet,

>>> print range(n, (m+1)*n, n)
[20, 40, 60, 80, 100] 

For Python3+

>>> print(list(range(n, (m+1)*n, n)))
[20, 40, 60, 80, 100] 

If you're trying to find the first count multiples of m, something like this would work:

def multiples(m, count):
    for i in range(count):
        print(i*m)

Alternatively, you could do this with range:

def multiples(m, count):
    for i in range(0,count*m,m):
        print(i)

Note that both of these start the multiples at 0 - if you wanted to instead start at m, you'd need to offset it by that much:

range(m,(count+1)*m,m)

Based on mathematical concepts, I understand that:

• all natural numbers that, divided by n, having 0 as remainder, are all multiples of n

Therefore, the following calculation also applies as a solution (multiples between 1 and 100):

>>> multiples_5 = [n for n in range(1, 101) if n % 5 == 0]
>>> multiples_5
[5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100]

For further reading:

• https://www.mathsisfun.com/definitions/natural-number.html
• https://www.mathwizz.com/arithmetic/help/help9.htm
• https://www.calculatorsoup.com/calculators/math/multiples.php

For the first ten multiples of 5, say

>>> [5*n for n in range(1,10+1)]
[5, 10, 15, 20, 25, 30, 35, 40, 45, 50]