And you know, I usually step these up in multiples of two until some maybe larger value this in multiples of two you

Only multiples of 10.

Let's do the multiples of 6

So non negative multiples of 3.

multiples back there.

Pay me back multiples.

So I circled all the multiples of 5.

Multiples of 6 6,12,18,24,30... i can keep going and the multiples of 8 are 8,16,24,. looks like we are done.

So, lets find the multiples of 15. You have

The other ones are some multiples of that length.

Those are literally the nearest multiples of ten that

Now that's the multiples of 15 up through 105. Obviously, we keep going from there. Now lets do the multiples of 6.

You can literally just go through the multiples of 4.

We could just look at the multiples.

So, all the multiples of three are coloured red, remainder of one will be coloured black and remainder of two will be coloured green.

We could pick one of these guys and keep taking their multiples, and then figuring out whether those multiples are both divisible by 5 and 15.

If you add up two multiples of three you always get another multiple of three( which is the sort of fact you use everday in math class)

So you take all the multiples of v1, you're going to get every point along this line.

So this is some weight on a, and then we can add up arbitrary multiples of b.

B goes straight up and down, so we can add up arbitrary multiples of b to that.

Let's do the multiples of 10. They are 10, 20, 30, 40... well, we already went far enough.

So let's we do this in a different color, so let me pick out the multiples of five.

One way is just to kind of look at the multiples of 9 and see if any of them are divisible by 12.

I have one group of two, two groups of two, three groups of two, four groups of two.

And that's just a fancy way of saying, look, if I'm on pass 17, I'm going to switch all the multiples of 17.

When you have two of anything two eyes, two ears, two noses, just two protrusions those all partake of the form of two.

Literally, just find and look at the multiples of each of the numbers... and then see what the smallest multiple is they have in common.

To see if we apply Bayes rule, which multiples the prior for spam ness with the conditional probability of each word given spam.

Three groups of two three times two.

Which means you take all of the multiples of v1, and all of the positions that they specify will create a set of points, and that is my line l.

two of clubs

two of diamonds

two of hearts

two of spades

Two of Diamonds

Two of them.

Two of them?

Lord of the Two Easts, Lord of the Two Wests,

Lord of the two Easts, Lord of the two Wests.

Two and it was two of spades, right?

Lord of the two Easts and Lord of the two Wests.

Lord of the two Easts, and Lord of the two Wests!

Lord of the two easts, and Lord of the two wests!

So that's two groups of 8 or two rows of 8

So to do that, lets think of the different multiples of 15, 6 and 10. and then find the smallest multiple, the least multiple they have in common.