I need to test whether each number from 1 to 1000 is a multiple of 3 or a multiple of 5. The way I thought I'd do this would be to divide the number by 3, and if the result is an integer then it would be a multiple of 3. Same with 5.
How do I test whether the number is an integer?
here is my current code:
n = 0
s = 0
while (n < 1001):
x = n/3
if isinstance(x, (int, long)):
print 'Multiple of 3!'
s = s + n
if False:
y = n/5
if isinstance(y, (int, long)):
s = s + n
print 'Number: '
print n
print 'Sum:'
print s
n = n + 1
a = 1400
a1 = 5
a2 = 3
b= str(a/a1)
b1 = str(a/a2)
c =b[(len(b)-2):len(b)]
c1 =b[(len(b1)-2):len(b1)]
if c == ".0":
print("yeah for 5!")
if c1 == ".0":
print("yeah for 3!")
Try this ...
public class Solution {
public static void main(String[] args) {
long t = 1000;
long sum = 0;
for(int i = 1; i<t; i++){
if(i%3 == 0 || i%5 == 0){
sum = sum + i;
}
}
System.out.println(sum);
}
}
You can use % operator to check divisiblity of a given number
The code to check whether given no. is divisible by 3 or 5 when no. less than 1000 is given below:
n=0
while n<1000:
if n%3==0 or n%5==0:
print n,'is multiple of 3 or 5'
n=n+1
This code appears to do what you are asking for.
for value in range(1,1000):
if value % 3 == 0 or value % 5 == 0:
print(value)
Or something like
for value in range(1,1000):
if value % 3 == 0 or value % 5 == 0:
some_list.append(value)
Or any number of things.
The simplest way is to test whether a number is an integer is int(x) == x
. Otherwise, what David Heffernan said.
For small numbers n%3 == 0
will be fine. For very large numbers I propose to calculate the cross sum first and then check if the cross sum is a multiple of 3:
def is_divisible_by_3(number):
if sum(map(int, str(number))) % 3 != 0:
my_bool = False
return my_bool
I had the same approach. Because I didn't understand how to use the module(%) operator.
6 % 3 = 0 *This means if you divide 6 by 3 you will not have a remainder, 3 is a factor of 6.
Now you have to relate it to your given problem.
if n % 3 == 0 *This is saying, if my number(n) is divisible by 3 leaving a 0 remainder.
Add your then(print, return) statement and continue your
You can simply use %
Modulus operator to check divisibility.
For example: n % 2 == 0
means n is exactly divisible by 2 and n % 2 != 0
means n is not exactly divisible by 2.
Source: Stackoverflow.com