I made a fizz-buzz-like algorithm with python, so I will post it.

When it is a multiple of 2, `~ is an even number`

When it is a multiple of 3, `~ is a multiple of 3`

When it is a multiple of 6, `~ is a multiple of 6`

Otherwise `~ is any other number`

Is displayed.

You can enter the number of judgments first.

First, create a function to determine the multiple.

```
def sample(x):
if (x % 3 == 0 and x % 2 ==0):
print(x,"Is a multiple of 6")
elif (x % 3 == 0):
print(x,"Is a multiple of 3")
elif (x % 2 == 0):
print(x,"Is even")
else:
print(x,"Is any other number")
```

A `:`

is added to the end of the ʻif` or ʻelse`

line,
ʻelif`, not elsif, Note that we are using ʻand`

instead of &&,
Other than that, it's not much different from ruby.

Next, be sure to enter the number of impressions.

```
print("How many do you want to display?")
y = int(input())
```

If it's just input, `y = input ()`

seems to be fine, but it wasn't judged as a number, so it's written like this.

Finally, make a multiple judgment for the number of times you enter.

```
for x in range(1, y + 1):
sample(x)
```

The number of executions is determined by `range (1, y + 1)`

.
Here, the process is to perform sample (x) for all integers from 1 to y entered earlier.
sample (x) is to call the function defined first.

The whole code is as follows.

```
# coding:utf-8
import sys
def sample(x):
if (x % 3 == 0 and x % 2 ==0):
print(x,"Is a multiple of 6")
elif (x % 3 == 0):
print(x,"Is a multiple of 3")
elif (x % 2 == 0):
print(x,"Is even")
else:
print(x,"Is any other number")
print("How many do you want to display?")
y = int(input())
for x in range(1, y + 1):
sample(x)
```

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