Learn how to Spherical as much as the Subsequent A number of of 5 in Python

The problem

Given an integer as enter, are you able to spherical it to the subsequent (that means, “larger”) a number of of 5?

Examples:

enter:    output:
0    ->   0
2    ->   5
3    ->   5
12   ->   15
21   ->   25
30   ->   30
-2   ->   0
-5   ->   -5
and many others.

Enter could also be any constructive or unfavorable integer (together with 0).

You may assume that each one inputs are legitimate integers.

The answer in Python code

Choice 1:

def round_to_next5(n):
    return n + (5 - n) % 5

Choice 2:

def round_to_next5(n):
    whereas npercent5!=0:
        n+=1
    return n

Choice 3:

import math
def round_to_next5(n):
    return math.ceil(n/5.0) * 5

Check instances to validate our resolution

inp = 0
out = round_to_next5(inp)
take a look at.assert_equals(out, 0, "Enter: {}".format(inp))

inp = 1
out = round_to_next5(inp)
take a look at.assert_equals(out, 5, "Enter: {}".format(inp))

inp = -1
out = round_to_next5(inp)
take a look at.assert_equals(out, 0, "Enter: {}".format(inp))

inp = 5
out = round_to_next5(inp)
take a look at.assert_equals(out, 5, "Enter: {}".format(inp))

inp = 7
out = round_to_next5(inp)
take a look at.assert_equals(out, 10, "Enter: {}".format(inp))

inp = 20
out = round_to_next5(inp)
take a look at.assert_equals(out, 20, "Enter: {}".format(inp))

inp = 39
out = round_to_next5(inp)
take a look at.assert_equals(out, 40, "Enter: {}".format(inp))