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iOS

Posted by Mike Cho
Mike Cho
570 Points

Swift Basic Course Extra Credit - Solution

Hello,

I know the following is annoyingly simple to most and probably embarrassing to ask...but here we go anyways:

In the extra credits sections the task was:

"Given a range of numbers from 1 to 100. Write a loop which prints out whether a number is a odd or even."

I wasn't sure whether I'm doing the right thing, so here is what I wrote:

let numbers = 1...100

for number in numbers {
    if (number % 2 == 0) {
        println("even")
    } else if (number % 2 != 0) {
        println("odd")
    }
}

In the console the output seems to be correct, but is there a better way to solve this?

The community is so overwhelmingly helpful here. Working hard to help others as well one day. Thank you!

Lester Cerda
Lester Cerda
Courses Plus Student 567 Points

Im with you Mike, i wasn't to sure either. This is my answer... It does show odds and even with numbers but i want the numbers to be first then odd or even eg.. 1 Odd not Odd 1.. any ideas???

import UIKit

for number in 1...100 { if number % 2 == 0{ println("Even (number)") } else { println("Odd (number)") }

}

7 Answers

Jeremy Hayden
Jeremy Hayden
1,740 Points

Mike, that was a very good answer, and it works. That is the main thing.
Just a of suggestion though.

Your output would be more meaningful if it let us know which numbers it was talking about when it says "odd" or "even". You could use a skill that you have already learned, string interpolation, to make the output say something similar to "3 is odd", "4 is even".

But other than that, your code is clean and logical.

Mike Cho
Mike Cho
570 Points

That's true! I will try that out now. Thank you =) !

Jeremy Hayden
Jeremy Hayden
1,740 Points
Jeremy Hayden
Jeremy Hayden
1,740 Points

Jacob Your code looks good as well. However your last else statement is not needed.

Let's go thru your code.

First your counting from 1-100.

Then your dividing each number by 2.

Let's skip 1 for a second and go directly to 2. 2 divided by 2 = 1. It divides perfectly, there is nothing left over. So your remainder is zero. So your first if statement is true, the remainder is zero, so it prints out 2 is even.

Next number is 3. 3 divided 2 is 1.5. Or 1 with a remainder of 1 that can be further divided to the .5

We don't care about further dividing to get the decimal. We just want to know if there is a remainder or not.

So

4 / 2 = 2 remainder 0

5 / 2 = 2 remainder 1

6 / 2 = 3 remainder 0

7 / 2 = 3 remainder 1

Ect. Even numbers divided by 2 will always have remainder 0.

Back to your code...

Let's say your next number is 3

3/2=1 remainder 1, or %=1

So the first if statement is false and it moved on

Next

3/1=3 with no remainder, or %=0 so this is true and it prints out 3 is odd

Well since whole numbers are either even or odd that means the last if statement will NEVER get executed because one of the 2 first ones always will be true. So the last if statement is useless.

Next, do you really even need the second if statement? I mean if your number is not even, can't we just assume it's odd?

So instead f an if statement, you could just do, else - print odd

Jake Webb
Jake Webb
13,028 Points

I really appreciate you breaking it down for me and helping me see things from a different perspective! Super Helpful!

Vasco Marques
Vasco Marques
938 Points

Very good explanation! Thanks Jeremy.

Jake Webb
Jake Webb
13,028 Points
Jake Webb
Jake Webb
13,028 Points

Here is my attempt..

for n in 1...100 {
    if n % 2 == 0{
        println("\(n) is Even")
    } else if n % 1 == 0 {
        println("\(n) is Odd")
    } else {
        println(n)
    }
}

Could someone explain my logic? I don't understand it but I'm received the desired output. I'm just not able to understand remainders for some reason.

Jason Hernandez Berland
Jason Hernandez Berland
1,575 Points

Sexy solution Jacob!

Looks like you understand very well from your code. Your conditional on lines 2 and 4 are checking for remainders:

if n % 2 == 0...
} else if n % 1 == 0 {...

So, is there a remainder when 1 % 2? Yep.

It does not satisfy == 0, so let's move on to the next condition...

is there a remainder when 1 % 1? No.

So it satisfies n % 1 == 0, and therefore prints output "1 is Odd"

ayushagarwal2
ayushagarwal2
982 Points
ayushagarwal2
ayushagarwal2
982 Points 
for num in 1...100{
    if (num%2==0)
    {
        println ("Even")
    }else
    {
        println ("Odd")
    }
}
Jonny Mack
Jonny Mack
497 Points
Jonny Mack
Jonny Mack
497 Points

Here is my attempt. I got the desired results, but I have a little bit different code than everyone else

for number in 1...100{ if (number % 2 == 0){ println("(number) is Even") } else { println("(number) is Odd") } }

kols
kols
27,007 Points
kols
kols
27,007 Points

My answer (I think this also should answer Lester's question above^^):

for i in 1...100 {
    if (i % 2 == 0) {
        println("\(i) is Even")
    } else {
        println("\(i) is Odd")
    }
}
alex3
alex3
5,387 Points
alex3
alex3
5,387 Points

This is my answer

for n in 1...100 {
    if n % 2 == 0 {
        println("\(n) is odd")
    } else if n % 3 == 0 {
        println("\(n) is even")
    } else {
        println(n)
    }
}