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.
We can use range() function in Python to store the multiples in a range. First we store the numbers till m multiples using range() function in an array, and then print the array with using (*a) which print the array without using loop. # of a number n without using loop. # (m * n)+1 incremented by n.
Use the modulo % operator to check if a number is a multiple of 10, e.g. if 100 % 10 == 0: . The modulo % operator returns the remainder from the division of the first number by the second. If the remainder is 0 , the number is a multiple of 10 .
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Finish the solution so that it returns the sum of all the multiples of 3 or 5 below the number passed in. Note: If the number is a multiple of both 3 and 5, only count it once.
Use the modulus operator: if (num % 5 == 0) //the number is a multiple of 5.
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)
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