I have one case like a=14 when the user enters 6 then it will be check as next largest number of the multiplication number.
if 6*2>14 so not working and then that 6*3>14 so it is working.
I have no idea what was the name of the math function.
I have one case like a=14 when the user enters 6 then it will be check as next largest number of the multiplication number.
if 6*2>14 so not working and then that 6*3>14 so it is working.
I have no idea what was the name of the math function.
I think you just want:
factor = ((a-1) // n + 1)
Where a
is you number like 14 and n is the smaller number to be multiplied by the factor
:
a = 15
n = 7
factor = ((a-1) // n + 1)
print("factor", factor)
# 3
print("multiplied:", n * ((a-1) // n + 1))
# 21
it's not clear from your question what happens with an exact match, but this will return the exact factor:
a = 14
n = 7
factor = ((a-1) // n + 1)
print("factor", factor)
# 2
print("multiplied:", n * ((a-1) // n + 1))
#4
This is my guess on what is meant based on the other comments:
user_number= int(input("What is your number? "))
check_number=int(input("What is the number to check against. "))
factor = 1
print(check_number, factor*user_number)
while(check_number>(factor*user_number)):
print(factor)
print("Not working.")
factor+=1
working_factor = check_number%user_number+1
print("using modulo: ", working_factor)
print("Working factor: ",factor)
import math
a = 14
b = 6
mutiple = math.floor(a/b)+1
value = mutiple*b
print (mutiple, value)
You can use math.ceil to calculate the ceiling of number/multiplier
and multiply with it multiplier
to get your result
import math
def func(number, multiplier):
return multiplier * math.ceil(number/multiplier)
The output will be
print(func(14,6))
#18
print(func(18,6))
#18
print(func(19,6))
#24
Example for number=14
and multiplier=6
, math.ceil(14/6)
gives 3
, and 6*3
gives 18