For this challenge, we'd like to know in a range of values from 1 to 100, how many numbers are both odd, and a multiple

Hi Walter,

In this challenge you are provided with a loop that iterates from 1 to 100 - at each iteration, the current value is held within the loop's variable, n. So, n starts out as 1, then 2, 3, etc.

You want to add the value of n into the results array (also provided for you), if the value of n at the time is both odd and a multiple of 7.

You can check for oddness by using the mod operator, % and the number 2. If a number is even number % 2 will equal zero. If the number is odd, that operation will generate a non-zero result. (It'll be 1, but let's just stick to non-zero, or not (!) zero). So that's the first test sorted, we want to see if n % 2 != 0, in other words, does the result of n % 2 generate a non-zero result.

Next we need to see if the value of n is a multiple of 7. We can use the same process. The mod operator, %, and the figure 7 will tell us if number divided by seven generates a remainder. If no remainder is generated, a zero result, then the number is a multiple of 7. That means our test is the same as with the odd test, but by using the value of 7 and checking for a zero result, not a non-zero one. So, that looks like n % 7 == 0.

We have our two tests. Both need to apply at the same time, so test 1 AND test 2, must be true for the current value of n to be added to the results array. We can AND two tests together using a double ampersand, &&.

This gives us the test:

if n % 2 != 0 && n % 7 == 0 {
  // add n to the array
}

OK so far?

Now the easy bit. How do we add an element to an array? We use the append() method and dot-notation. The array is called results and the parameter we pass into append() is n.

That gives us:

if n % 2 != 0 && n % 7 == 0 {
  // add n to the array
  results.append(n)
}

Let me know how you get on.

Steve.