# Adding Large Numbers

Hello! Today we’re going to take a look at how to **add large numbers**. We do simple addition in our heads. \(5+6=11\), and we don’t even really have to think about it. But when numbers get bigger, we need to follow a system to add them. For instance, what if we took a look at the problem \(4,537+692\) ? This one we can’t do in our heads, so let’s take a look at how we would solve this.

The key to adding up numbers is to line them up by their place values, stacking them on top of one another. So, for this problem, that would look like this. We have the first number 4,537, and then we want to line up 692 underneath by its **place value**. So that would go right here. And then we’re adding, so this is what it would look like.

\(4537\) \(+\)\(692\)

Notice that the ones placed in the first number should be right above the ones place in the second number. The tens place is above the tens place, the hundreds place is above the hundreds place. And then we have a thousands place in our top number but not in our bottom number, so that’s okay. So, now what we want to do is add from right to left, our ones places, and then our tens places, and so on. So let’s start doing that. We have \(7+2=9\), so we’re going to write a 9 right here.

\(4537\) \(+\)\(692\) \(9\)

Now let’s add \(3+9\), well \(3+9=12\), and we can’t fit 12 into this one digit place right here. So what we’re going to do is, we’re going to write down the 2 part of the 12 and then the 10 part — the 1, is going to be carried to the next place value, just like this.

\(\phantom{0}1 \phantom{00}\) \(4537\) \(+\)\(692\) \(29\)

So, it looks a little bit weird, but when we think about it the 1 in the number 12 is really a 10, and because each time we move up one place value we increase by a tens place, we can carry the 1 and it’s the same thing. So now we’re going to add this next column. \(6+5=11\), and then we’re going to add this 1 also, so \(11+1=12\). Again, we’re going to do the exact same thing, so we’re going to write down the 2 and carry our 1.

\(11\phantom{00}\) \(4537\) \(+\)\(692\) \(229\)

There’s nothing under the 4, which means we’re just going to pretend it’s a 0, we don’t have to add anything. And then we’ll just move to the 4 and then add the 1 that we carry, so \(4+1=5\).

\(11\phantom{00}\) \(4537\) \(+\)\(692\) \(5229\)

So \(4,537+692=5,226\).

Now that we’ve figured this out, let’s look at another problem so that we can get some more practice under our belts.

What if we have an even larger number, like \(45,982+9,673\) ?

The good news is, we can use the exact same system that we did last time to solve this problem. So again, we’re going to line up our numbers by their place value.

\(45982\) \(+ \text{ }9673\) \(\)

And then we’re going to add from right to left.

\(45982\) \(+\text{ }9673\) \(5\)

\(8+7=15\). Remember, since we have a two-digit number, we write down the 5 (the second part of it) and then we carry the 1 from the 15 (that’s the number that was in our tens place).

\(\phantom{0}1\phantom{00}\) \(45982\) \(+\text{ }9673\) \(55\)

So we carry it over and then add this whole column: \(9+6=15+1=16\). We’re gonna do the same thing with carrying, so write down the 6 and carry the 1.

\(11\phantom{00}\) \(45982\) \(+\text{ }9673\) \(655\)

\(9+5=14+1=15\), so we’re gonna write down a 5, carry a 1.

\(111\phantom{0}\) \(45982\) \(+\text{ }9673\) \(5655\)

Remember, there’s nothing here so we can pretend it’s a 0. \(0+4=4+1=5\), so we write down a 5.

\(111\phantom{0}\) \(45982\) \(+\text{ }9673\) \(55655\)

And now we can see that \(45,982+9,673=55,655\).

I hope that this video on adding large numbers made it a little bit clearer for you. Thanks for watching, and happy studying!