If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

It has always been fun for me to play with mathematic problems and when it comes to the mathematics we should check Project Euler.

Here are two solutions I have found for the first problem:

The first option would be like this:

```
$sum = 0;
for ($i = 0; $i < 1000; $i++) {
if ($i % 3 == 0)
{
$sum += $i;
continue;
}
if ($i % 5 == 0)
{
$sum += $i;
}
}
echo $sum;
```

For every number which is lower than a thousand we check if this given number divisible to 3 *or* 5.

The point here is not to use the `else`

statement.

PHP provides a `continue`

statement to skip the remaining part of the current loop. In our case, if the given number is divisible by 3 we skip the second if statement.

And the second option would be like this:

```
$sum = 0;
for ($i = 0; $i < 1000; $i++) {
foreach ([3, 5] as $divisor)
{
if ($i % $divisor == 0)
{
$sum += $i;
break;
}
}
}
echo $sum;
```

We provide the numbers to be the divisor as a variable. (`$number`

) While we count from 0 to 1000. We check for every divisor and if it divisible by the given `$number`

we are good to break the `foreach`

loop which will let us skip the next numbers in the queue.

**Execution time:** 0.0009s `(both ways)`