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## 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))
```

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