Recently came across this technique of multiplying two $2$-digit numbers involving the same digit at the tens place and the sum of digits at units place being $10$. E.g., $73 X 77$ has same digit at the tenth place, viz., $7$ and the sum of digits at the ones place is $7+3 = 10$.
As per the technique the easy way to solve this simply multiply the digits at ones place together which in the above case would be $7. 3 = 21$ and the digit at the tenth place by the next number, i.e., $7.8=56$ in the above. The final answer then becomes after concatenating the two answers thus obtained, and hence $73.77=5621$. Similarly, $68.62=4216$ obtained by multiplying $6.7=42$ and $8.2=16$ and concatenating the two to obtain $4216$.
Can anyone explain the math behind this seemingly simple math trick?
$(10x + a)(10x +b) = 10x(10x+a+b) + ab$. Furthermore, if $a+b = 10$ , then this simplifies to $10x(10x + 10) + ab = 100x(x+1)+ab$.
That is, what we are doing, is basically the following: given say $72 \times 78$, we get $x(x+1) = 56$, and $ab = 16$, so the answer is $5616$. The reason why you can just join $56$ and $16$ is because $ab$ is always a two digit number, and $100x(x+1)$ has it's last two digits as $00$, so linear addition becomes concatenation.
By the way, this assumes nothing about $x$ at all. For example: $$ 173 \times 177 = (17 \times 18) *(3 \times 7) = (306)*(21) = 30621 $$
So $x$ could be anything. All we require is that $a+b = 10$ and all but the last digit of the numbers are the same.
A special case arises when $a=b=5$ for then $ab=25$, and essentially you are squaring the number $(x5)^2$. For example, $75^2 = (7 \times 8)*25 = 5625$.