Wednesday, January 13, 2010
To multiply a number by 11 and add enough pairs of numbers from right within the given number except the numbers on the edges that should be repeated.
For example:
consider the following product: 324 x 11
* From the right type 4 (the last number to the right of 324) first digit of the product;
* Try the sum 4 +2 = 6 (the sum of 1 'and 2' figure), you get the second digit of the product;
* Try the sum 3 +2 = 5 (the sum of the 2 'and 3' figure), you get the second digit of the product;
* Repeat 3, last number of the product.
Summing up (*):
(3) (3 +2) (4 +2) (4) = 3564
(*) Groups are not intended parentheses multiplied together
Now let's see what happens when the sum of two numbers exceeds the internal ten.
Consider the product:
967 x 11
* Begin to write from right to left, the first digit of the product is the 7;
* Sum 6 +7 = 13 write 3 (2 'figure of the product) and over 1;
9 +6 = 15 * sum added to the carryover of the amount above 15: 15 +1 = 16 writing 6 with carryover of 1;
* Last figure 9 we add the carry of the sum above 9 +1 = 10 which is the last digit of the product;
and therefore:
967 x 11 = 10637
Simple right?
With a little 'training can become very fast and impress your classmates :-)
1 Comment:
Whuaaah, it's so nice tips! Math will be the easiest subject in school if more tips as here. Thanks for your sharing. :)
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