# Daily Challenge #25

I hope you still enjoy the daily challenges. Like i said before if you want you can also make a daily challenge. The challenge is then also to do it before i do.

## Oddish vs. Evenish

Create a procedure that determines whether a number is Oddish or Evenish . A number is Oddish if the sum of all of its digits is odd, and a number is Evenish if the sum of all of its digits is even. If a function is Oddish , return `"Oddish"` . Otherwise, return `"Evenish"` .

For example, `oddishOrEvenish(121)` should return `"Evenish"` , since 1 + 2 + 1 = 4. `oddishOrEvenish(41)` should return `"Oddish"` , since 4 + 1 = 5.

### Examples

``````oddishOrEvenish(43) ➞ "Oddish"

oddishOrEvenish(373) ➞ "Oddish"

oddishOrEvenish(4433) ➞ "Evenish"``````
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Here it is:

And for this:

oddishOrEvenish(43) ➞ “Oddish”
oddishOrEvenish(373) ➞ “Oddish”
oddishOrEvenish(4433) ➞ “Evenish”

Here is a solution:

``````    public String OddishOrEvenish(int num){

int sum = 0;

String str = String.valueOf(num);

String[] n = str.split("(?<=.)");

for ( String i:n) {
sum += Integer.parseInt(i);
}
String result = "Oddish";
if(sum%2==0){
result = "Evenish";
}
return result;
}
``````

Blockly saves so much time and labour

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Thank you

For extension developers

We can make 25 new extensions one from each challenge or a single extension having 25 blocks

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(remainder of)
I am not finding this block in Kodular…

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Search for ‘modulo of’

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