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## Multiplication by 10s, 100s, and 1000s

# Multiplying 10s

CCSS.Math:

## Video transcript

- [Voiceover] 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 computations to do. And there's gotta 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 gonna break up 40
and say, instead of 40, four times 10. Four times 10 and 40 are equivalent. They're the same thing,
so I can replace the 40 with a four times 10. And then for my 70, same thing. I can break this up and
write seven times 10. Seven times 10. So these two expressions, 40 times 70 and four times 10 times seven times 10, 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 10s. So I'll solve this one, knowing that I'll get the same
solution as I would have for this top expression. So what we can do is we
can re-order 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 five times two, for example, that would be the same as two times five. They're both 10. Five twos or two fives,
either way, it's 10. 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 gonna put my
one-digit numbers first. Four times seven. And then, I'll put the
two-digit numbers, the 10s, times 10 and the other times 10. So we have all the same factors, all the same numbers, in
both of these expressions. They've just been re-ordered. And now, I'll solve going across. Four times seven is 28. And now 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 zero to the end. One zero for that zero in 10, because 28 times 10 is 28 10s, 28 10s, 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
10s, we add two zeros. So, 28 times 10 times 10, is 2,800. Which means that this
original expression we had, 40 times 70, also has a solution of 2,800, or two thousand, eight hundred. Let's try another example where we're multiplying 10s like this. Let's try, let's do
something like, let's say, maybe 90 times, how about, 30. 90 times 30. So the first thing I'm gonna
do is break up these numbers so that I have 10s, because again, for me 10s are easier to multiply
than numbers like 90 and 30. So for 90, I'll write nine times 10, and for 30 I'll write three times 10. The expressions are equivalent. We've just written it in another way. And now, I'll re-order these numbers to put the one-digit numbers first. So, nine times three, and then I'll put the
10s: times 10, times 10. Because we need to have
all the same numbers, even if we change the order. So we have the nine, three, the first 10 and the second 10. And now, finally, we multiply. Nine times three is 27. 27 times 10 will be 27 10s, or 27 with a zero on the end. And 270 times 10 will be 270 10s, or 270 with a zero on the end. So going back to the original question, 90 times 30 is equal to 2,700. Or two thousand, seven hundred.