it works for everything but it uses something called base reference numbers (sounds complicated but the book explains them and it becomes easy - oh and i'm no genious!).
e.g in this example we are using two base reference numbers 20 & 80 and as 20 goes in to 80 four times we can substitue 80 to a 4.)
23 x 87
BR (20 x 4) 23 x 87
both numbers are higher than the reference numbers (i.e 23 is higher than 20 and 87 is higher than 80) so we draw circles above
+3 +7
(20 x 4) 23 x 87=?
now multiply the 3 above the 23 by multiplication factor 4. (3 x 4 = 12
so it now looks like this
+12
+3 +7
(20 x 4) 23 x 87=?
now add 12 to the 87 which gives 99
now multiply 99 by the other base reference number 20
99 X 20 = 1980
(99 x 20 is easy because you do 99 + 99 which gives 198 and then just add a 0 to signify the tenths.)
and then multiply 3 by the 7
99 X 20 = 1980
3 x 7 = 21
= 2001
add them together and done!
so the problem is this
+12
+3 +7
(20 x 4) 23 x 87 = 99
=1980
= 21
=2001
looks a lot to take in but the book is an easy read and explains it far better.....
hope this helps
oh I forgot!
If both numbers in the circles are below the reference number (98 is -2 from a hundred) then when you multiply the numbers in the circles the result you subtract + & - = -
If both numbers in the circles are above the reference number (102 is +2 from a hundred) then when you multiply the numbers in the circles result you add+ & + = +
If one number is above and one is below the reference number (102 is +2 from a hundred but 98 is -2 below) then when you multiply the numbers in the circles the result you add- & - = +