To check whether a number is multiple of second number

Question

I want to check whether a number is multiple of second. What's wrong with the following code?

def is_multiple(x,y):
    if x!=0 & (y%x)==0 :
       print("true")
    else:
       print("false")
    end
print("A program in python")
x=input("enter a number :")
y=input("enter its multiple :")
is_multiple(x,y)

error:

TypeError: not all arguments converted during string formatting

Solution 1

You are using the binary AND operator &; you want the boolean AND operator here, and:

x and (y % x) == 0

Next, you want to get your inputs converted to integers:

x = int(input("enter a number :"))
y = int(input("enter its multiple :"))

You'll get a NameError for that end expression on a line, drop that altogether, Python doesn't need those.

You can test for just x; in a boolean context such as an if statement, a number is considered to be false if 0:

if x and y % x == 0:

Your function is_multiple() should probably just return a boolean; leave printing to the part of the program doing all the other input/output:

def is_multiple(x, y):
    return x and (y % x) == 0

print("A program in python")
x = int(input("enter a number :"))
y = int(input("enter its multiple :"))
if is_multiple(x, y):
    print("true")
else:
    print("false")

That last part could simplified if using a conditional expression:

print("A program in python")
x = int(input("enter a number :"))
y = int(input("enter its multiple :"))
print("true" if is_multiple(x, y) else "false")

Solution 2

Some things to mention:

1. Conditions with and, not & (binary operator)
2. Convert input to numbers (for example using int()) - you might also want to catch if something other than a number is entered

This should work:

def is_multiple(x,y):
    if x != 0 and y%x == 0:
        print("true")
    else:
        print("false")

print("A program in python")
x = int(input("enter a number :"))
y = int(input("enter its multiple :"))
is_multiple(x, y)