 # An obvious trick (combo) 2 digits with last 2 numbers adding to 10

Most everyone knows that if you have

A square ending in 5 ala, 25, 125, 35 1205

\begin{align} x &= (a5)^2 \\ &= a * (a+1) + 25 \\ \\ 625 &= 25^2\\ &= (20 * 30) + 25\\ &= 625\\ \end{align}\\

Generalizing this when you have the last digit of a number summing to 10 that you can simply do the same 21*29, 37*33. (If you have 2 digit squares memorized then you obviously just subtract from the square and save yourself further calculation by using the fact ( I generally see 21 * 29 immediately as 625 - (8/2)^2 and struggle with the complement to get 609, but ignore that for the moment)

ab * ac \text{ where } b+c =10\\ \begin{align} x &= a * (a+1) + (b*c)\\ \\ 31 * 39 &= (30 * 40) + ( 1 * 9 )\\ &=1209 \end{align}

Now the obvious but fun part I haven’t seen tossed out before; 23 * 37
Yup… you can still quickly use the symmetry, numbers gotta love em.

|ab| * |(a+j)c| \text{ where } b+c = 10 \\
\begin{align} x &= ab * (a+j)c\\ &= (a * (a+1) + b * c ) + ( ab * j )\\ \end{align}
\begin{align} 29 * 31 &= (29 * 21) + (29 * 10)\\ &= ((20 * 30) + 9) + (29 * 10)\\ & = 609 + 290\\ &= 899\\ \end{align}

Of course, you will point out that if have your squares nailed then
900-1=899 \$ lol.
and even if the number was
21 * 39
then you are still centred on 30 and
900 - 81 = 819

but 34 * 76 might feel like a bit of a pain in the butt so it is nice to have a couple of choices when you see this. ( although 55^2 - 21^2 isn’t that hard )

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