Quick multiplication How can we explain this with equations ?
Quick Multiplication of any two numbers whose last two digits add up to 10 and all other numbers are the same
$32\times 38 = 3\times (3+1)|(2\times 8)\implies  (3\times 4)|16\implies 12|16 \implies 1216$
$81\times 89 \implies 8\times(8+1)|(1\times9)
\implies(8\times9)|09\implies72 | 09
\implies 7209$
$124 \times 126 \implies 12\times(12+1) | (4\times6)
\implies(12\times13)|24\implies156|24
=15624$
 A: $(10 a + b)(10 a + d) = 100 a^2 + 10 a (b + d) + b d$
Since $b + d = 10$, that's the same as $100(a^2 + a) + bd = 100a(a+1) + b d$.
A: Express the numbers as 10*x + y and 10*x + (10-y).
Here x is the part of the number other than last digit,  y and 10-y are the last digits.
So in case of 32 * 38, x=3, y=2
in case of 124 * 126, x=12, y=4 etc.
Now multiply the expressions for the 2 numbers i.e.
(10*x + y) * (10*x + (10-y))
You will get
100*x*x+100*x+a*(10-a)
which after simplification becomes
100*x*(x+1)+a*(10-a)
The first part is a number ending in 2 zeros as it has a factor of 100.
the last part is nothing but the multiplication of last digits of the original pair of numbers.
A: So the two numbers are $ 10d + a $ and $ 10d + 10 - a $, where is the last digit, necessarily non-zero, and $ d $ is the number formed by the rest of the digits. Hence $$ (10d + a)(10d + 10 - a) = 100(d^2 + d) + a(10 -a) $$ The number $ a(10 -a) $ is either one or two digits, so $ 100(d^2 + d) $ appears to the left of $ a(10 -a) $ (including a zero in the case of $ 1 \cdot 9 $. 
