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Multiplying 10s | Math | 4th grade | Khan Academy


3m read
·Nov 11, 2024

Let's multiply 40 times 70.

So, 40 times we have the number 70.

So, we could actually list that out, the number 70, 40 different times and add it up, but that's clearly a lot of computation to do, and there's got to be a faster way.

So, another way is to stick with multiplication but see if we can break these numbers up, this 40 and this 70.

Decompose them, break them up in some way to get numbers that might be a little easier to multiply with.

For me, multiplying by 10 is the easiest number because I know the pattern to add a zero.

So, I'm going to break up 40 and say instead of forty, four times ten.

Four times ten and forty are equivalent; they're the same thing.

So, I can replace the forty with a four times ten.

And then, for my seventy, same thing.

I can break this up and write seven times ten, seven times ten.

So, these two expressions, forty times seventy and four times ten times seven times ten, are equal; they're equivalent.

So, they'll have the same solution, but for me, this one down here is simpler to work out because of these times tens.

So, I'll solve this one knowing that I'll get the same solution as I would have for this top expression.

What we can do is we can reorder these numbers in a different order to again continue making this question easier for us to solve.

Because in multiplication, the order doesn't matter.

If we have 5 times two, for example, that would be the same as two times five; they're both ten.

Five twos or two fives, either way, it's ten.

So, we can change the order of the numbers without changing the answer.

So again, we're going to change our expression a little bit, but what we're not going to change is the solution.

So, I'm going to put my one-digit numbers first: 4 times 7, and then I'll put the two-digit numbers, the tens times ten and the other times ten.

So, we have all the same factors, all the same numbers in both of these expressions; they've just been reordered.

And now I'll solve going across.

4 times 7 is 28, and then we have 28 times 10 and times another 10.

Well, the pattern for times 10 that we know is when we multiply a whole number like 28 times 10, we will add a 0 to the end.

One 0 for that 0 in 10 because 28 times 10 is 28 tens.

28 tens or 280.

And that multiplied, 28 times 10.

And then if we multiply by this other 10, well, we have to add another zero.

Multiplying by 10 adds a zero, so if we multiply by two tens, we add two zeros.

So, 28 times 10 times 10 is 2800, which means that this original expression we had, forty times seventy, also has a solution of twenty-eight hundred or two thousand eight hundred.

Let's try another example where we're multiplying tens like this.

Let's try, let's do something like, let's say maybe 90 times about 30.

90 times 30.

So, the first thing I'm going to do is break up these numbers so that I have tens, because again for me, tens are easier to multiply the numbers like 90 and 30.

So, for 90 I'll write 9 times 10, and for 30 I'll write 3 times 10.

The expressions are equivalent; we've just written it in another way.

And now I'll reorder these numbers to put the one-digit numbers first: so nine times three, and then I'll put the tens times 10 times 10, because we need to have all the same numbers, even if we change the order.

So we have the 9, 3, the first 10, and the second 10.

And now finally, we multiply.

9 times 3 is 27.

27 times 10 will be 27 tens or 27 with a zero on the end, and 270 times 10 will be 270 tens or 270 with a zero on the end.

So, going back to the original question, 90 times 30 is equal to 27 hundred or 2,700.

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