在 Python 中查找数字的倍数

Question

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.

Answer 1

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)

Answer 2

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] 

Answer 3

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

Answer 4

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]

Answer 5

You can do:

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

Answer 6

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])

Answer 7

Another method that can be done is trying to make a list. Here's my example for getting the first 20 multiples of 7.

Input:

multiples_7 = [x * 7 for x in range(1,21)] 

print(multiples_7)

Output:

[7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133, 140]

Answer 8

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) ] 

Answer 9

For first 10 multiples of 5 you can do as

import numpy as np
#np.arange(1,11) array from 1 to 10 
array_multipleof5 = [5*n for n in np.arange(1,11)]
array_multipleof5 = np.array(array_multipleof5)
print(array_multipleof5)

Answer 10

How to calculate the first n multiples of a given number x, in the compact python's lambda notation

n_multiples_of_x = lambda n,x : list( range(x, x*n + 1, x) )

Tests:

assert n_multiples_of_x(5, 5) == [5, 10, 15, 20, 25]