Lesson 10

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Did you notice that when you had multiplicands with long strings of 3s in them, that the products would have long strings of the same digit? It might have been strings of 9s or 0s, depending on the digit immediately to the right of the string of 3s.

The same thing goes for long strings of any digit. Let’s try some:

633,331*3=1,899,993
533,336*3= 1,600,008
2,333,339*3=7,000,017

Notice that in that last example, there was a 1 at the end of a string of 0s in the product. that is because when we got to the 7 in the multiplicand, it forced us to carry a 2 (because 7*3=21), turning the 9 in the tens-columm of the product into an 11, instead of a 10.

Let’s try a string of 5s:
955,551=2,866,653
355,554=1,066,662
255,558=766,674

Did you notice the the digits in bold are different? The final number at the end of the string of similar numbers will always be determined by the number carried (or not carried) to it.

A word to the wise - at some point, if you get really good at multiplication, some wise-guy is going to think he will mess you up by throwing a string of 9s at you. For example, if you tell him to give you a large number to multiply, like in the millions, inevitably some jerk is going to come up with 9,999,999.

You’ll be ready for it, though, if you realize that the first digit of the product will be whatever digit you are multiplying the string of nines by (that’s the multiplier) minus 1. The last digit of the product, will be nine minus that digit. All the other digits in the multiplicand will be 9s.

So there’s practically no work. When he throws 9,999,999*3 at you, you can immediately throw back, “Twenty-nine million, nine hundred ninety-nine thousands, nine hundred ninety-seven” (because 7 is 9-2).

If the problem had been 9,999,999*9, the product would be 89,999,991.

Had it been 9,999,999*7, the answer would be 39,999,993

And so on.

If you work this problem out with the way we have learned, you will immediately see why it must work.

A way to look at this is just as if the multiplicand had been a one-digit number. You may have at some time noticed that any one-digit number times nine had a digit which is one less than the multiplier in the tens-column, and nine minus that digit in the ones-column.

9*1= 9 (which is the same as 09)
9*2=18
9*3=27
9*4=36
9*5=45
9*6=54
9*7=63
9*8=72
9*9=81

So when you multiply, say, 999*7, you basically pretend you are multiplying 9*7, then put a string of 9s in the middle of the answer. How many 9s? It’s easy: you know that the answer to the above will be in the thousands. You know the first and lasts digits are 6 and 3, so just fill in the hundreds- and tens-columns with 9s.

The rule of thumb for the number of 9s in the string is: Write one less nine than was in the multiplicand.

In a nutshell, the basic rule for multiplying a number made up of a string of nines by any one-digit number is:
Multiply nine times the digit, then write one less nine than was in the multiplicand between the digits.

Final example:
999,999*8= 72 with five 9s in the middle.
Thats 7,999,992.

Hotcha!

There’s no practice sheet for this lesson. Here are some for you to practice, though:

999*7=

99,999*9=

9*444,4444

9*111,111=

2,222*9=

9*333,333=

By now you should have this concept “locked in”.

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