I must test whether every number from 1 come 1000 is a multiple of 3 or a lot of of 5. The method I thought I"d carry out this would certainly be to divide the number through 3, and if the result is an integer climate it would be a lot of of 3. Exact same with 5.

How execute I check whether the number is one integer?

here is my current code:

n = 0s = 0while (n

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You do this utilizing the modulus operator, %

n % k == 0evaluates true if and also only if n is precise multiple that k. In primary school maths this is well-known as the remainder native a division.

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In her current approach you execute a department and the an outcome will be either

always an essence if you use integer division, oralways a float if you use floating suggest division.

It"s just the wrong way to go about testing divisibility.


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You can simply usage % Modulus operator to inspect divisibility.For example: n % 2 == 0 means n is exactly divisible through 2 and also n % 2 != 0 means n is not precisely divisible by 2.


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You have the right to use % operator to examine divisiblity the a given number

The code to check whether offered no. Is divisible by 3 or 5 when no. Much less than 1000 is provided below:

n=0while n
This code appears to execute what you space asking for.

for value in range(1,1000): if value % 3 == 0 or worth % 5 == 0: print(value)Or other like

for value in range(1,1000): if value % 3 == 0 or worth % 5 == 0: some_list.append(value)Or any number of things.


I had the exact same approach. Since I didn"t understand just how to usage the module(%) operator.

6 % 3 = 0*This means if you division 6 by 3 you will certainly not have actually a remainder, 3 is a element of 6.

Now you have to relate it to your offered problem.

if n % 3 == 0*This is saying, if mine number(n) is divisible by 3 leaving a 0 remainder.

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Add her then(print, return) statement and continue your


a = 1400a1 = 5a2 = 3b= 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!")
For little numbers n%3 == 0 will certainly be fine. Because that very big numbers ns propose to calculate the cross sum an initial 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
Try this ...

public course Solution { public revolution void main(String<> args) { lengthy t = 1000; long sum = 0; for(int i = 1; ns
The simplest method is to check whether a number is an integer is int(x) == x. Otherwise, what David Heffernan said.


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ask the user to kind an integer and also then prints "Yes" if that integer is divisible through 3, otherwise prints "No" (Python)