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# Using associative property to simplify multiplication

Sal uses the associative property to multiply 2-digit numbers by 1-digit numbers.

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• Can't we do like that 3*21 = (3*20)+(3*1)=60+3=63
For me it is much more easier to do them
like 83 *4 =(80*4)+(3*4)=320+12=332? • i dosent know how to do this very well can i has some help? • There was so many numbers and number sentences I couldn't understand anything. Why are we doing so much things? • • Does it matter, if you wrote the answer to be 15 x 3 = (3x3) x 5 = 9 x 5 = 45. / video time. • • • I am very confused I realy need some help.If you help i will give you an up vote.Thank you. • • • Okay. So, when multiplying a value of two digits or longer, by a value of one digit, you are making a one-digit by one-digit multiplication at each place value, but since they are of higher place values, each also needs to have their appropriate place values.
The video has five times 18. So first you do 5 times 8. That equals 40. Then you have a 5x1, which equals 5. But since the 1 is in the ten's place, the value of this product is 50.
After you finish the multiplication of the individual digits, you add them according to their place values. Previously, you were pretending that every value was of the one'a place, but for the correct product you will now need to add each to their actual place values. Both the 5 and 4 are in the ten's place, and is why there is 90.

Sometimes, you may have a digit on a line that, when added to digit below, gives a sum greater than 10. In this case, the next higher place value needs to be increased by the  so the 10s are not forgotten (that is what occurred when the 5 times 8 was done).
For example, if you have 27x9, you first multiply the 7 by the 9 for a 63. Write a 3 in the one's place, and a 6 above the digits in the ten's place so it will be added later.
Next, you multiply the 2 by the 9. This equals 18, so write an 8 in the ten's place and a 1 in the next higher place value.
Now, you add the 6 with the 8 and get a 14. The 4 goes into a row in the ten's place, and the 1 goes to the next higher place value.
Since there are two 1s, you get a 2. These are in the hundreds place, so there are two hundreds (for the hundreds themselves), and a final value 243.

The long multiplication is done this way. As I said before, at first you are doing a one-digit by one-digit multiplication at each step, which should not be difficult if you have memorized your one-digit times tables. So as you do the multiplication you are lining them up to their respective place values in advance.

Here is an example when there are higher place values for both digits:
If you have something like a two-digit multiplied by a two-digit, such as 14x32, when you multiply the 3 in the ten's place by each single digit, those are multiples of 10. And when you do the 3x1, both are in the ten's place, so the 3 gets multiplied by two 10s for a multiple of 100 in the final value.

Hope this helps in better understanding and clears things up.