[python] finding multiples of a number in Python

I'm trying to write a code that lets me find the first few multiples of a number. This is one of my attempts:

def printMultiples(n, m):
for m in (n,m):
    print(n, end = ' ')

I figured out that, by putting for m in (n, m):, it would run through the loop for whatever number was m.

def printMultiples(n, m):
'takes n and m as integers and finds all first m multiples of n'
for m in (n,m):
    if n % 2 == 0:
        while n < 0:
            print(n)

After multiple searches, I was only able to find a sample code in java, so I tried to translate that into python, but I didn't get any results. I have a feeling I should be using the range() function somewhere in this, but I have no idea where.

This question is related to python math range

The answer is


def multiples(n,m,starting_from=1,increment_by=1):
    """
    # Where n is the number 10 and m is the number 2 from your example. 
    # In case you want to print the multiples starting from some other number other than 1 then you could use the starting_from parameter
    # In case you want to print every 2nd multiple or every 3rd multiple you could change the increment_by 
    """
    print [ n*x for x in range(starting_from,m+1,increment_by) ] 

You can do:

def mul_table(n,i=1):
    print(n*i)
    if i !=10:
        mul_table(n,i+1)
mul_table(7)

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 this is what you are looking for -

To find all the multiples between a given number and a limit

def find_multiples(integer, limit):
    return list(range(integer,limit+1, integer))

This should return -

Test.assert_equals(find_multiples(5, 25), [5, 10, 15, 20, 25])

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:


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]

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