how to code this..

in 200 to 1

i need to decrease by 3 and displaying odd or even numbers..

can you help me..

ty :)

majimboo04
0
Newbie Poster

how to code this..

in 200 to 1

i need to decrease by 3 and displaying odd or even numbers..

can you help me..

ty :)

Jump to PostYou need two things to do this. Firstly, you need to make a

`for`

loop that runs over all the numbers. Secondly, you need to use the`%`

operator on each.`%`

gives you the "remainder" of an integer division. If you divide an integer by 2 …

prakash89.gitam
-3
Newbie Poster

As i understood from your question is that, you want numbers to be displayed even or oddthe number sequence will be from 200 to 1 by decreasing by 3

```
c=0;
for(int i=200;i>0;i=i-3)
{
if(c)
{
cout<<i+" is even";
c=1;
}
else
{
cout<<i+" is odd";
c=0;
}
}
```

Simple logic... as we know if u decrease it by 3 alternate numbers will be even, so i did like this.

You can also do by chceking if it is divisible by 2 or not..

Correct me if i am wrong.

WaltP
commented:
We dont solve homework problems for others. We HELP them find their own answers
-3

prakash89.gitam
-3
Newbie Poster

Sorry small change in code...

```
c=0;
for(int i=200;i>0;i=i-3)
{
if(c)
{
cout<<i+" is odd";
c=0;
}
else
{
cout<<i+" is even";
c=1;
}
}
```

WaltP
commented:
And again
+0

ravenous
266
Posting Pro in Training

You need two things to do this. Firstly, you need to make a `for`

loop that runs over all the numbers. Secondly, you need to use the `%`

operator on each. `%`

gives you the "remainder" of an integer division. If you divide an integer by 2 and look at the remainder, it will be 0 for even numbers and 1 for odd numbers:

```
x % 2 == 0; // if x is even
y % 2 == 1; // if y is odd
```

Have a play with those ideas and see how you get on. Post your code if you have any further problems

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