Calculator-Killer-3

First, $A^* = A^2 - A +1$

Then, say $B = A - 1$

Then, $B^* = A^2 - 3A +3$

Thus, $A^* - (A-1)^* = 2(A-1) = 2 B$

So, $10^*-9^*+8^*-7^*+6^*-5^*+4^*-3^*+2^*-1^*$

is equal to $2(9 + 7 + 5 + 3 + 1) = 2 \times 25 = 50$

$A^*=A\div A + A \times A - A=1+A(A-1) \\ 10^* - 9^* + 8^* -7^* + 6^* - 5^* + 4^* -3^* +2^*-1^* \\ 1+10\times9-(1+9\times8)+1+8\times7-(1+7\times6)+1+6\times5-(1+5\times4)+1+4\times3-(1+3\times2)+1+2\times1-(1+1\times0) \\ \cancel{1}+10\times9- \cancel{1}-9\times8+\cancel{1}+8\times7-\cancel{1}-7\times6+\cancel{1}+6\times5-\cancel{1}-5\times4+\cancel{1}+4\times3-\cancel{1}-3\times2+\cancel{1}+2\times1-\cancel{1}-1\times0 \\ 10\times 9 -9\times8+8\times 7-7\times6+6\times5-5\times4+4\times3-3\times2+2\times1-1\times0 \\ 9(10-8)+7(8-6)+5(6-4)+3(4-2)+1(2-0)=18+14+10+6+2=\boxed{50}$

One can simplify A* to equal $1 + A^2 - A$ We now have the equation: $(1 + 10^2 - 10) - (1 + 9^2 - 9) +... + (1 + 2^2 - 2) - (1 + 1^2 - 1)=?$

The constant 1 is both added and subtracted 5 times, so in the end, it cancels itself out. Because it is negligible, a modified equation could be:

(10^2 - 10) - (9^2 - 9) +... + (2^2 - 2) - (1^2 - 1)=?]

We can now rearrange this to be:

$(10^2 - 9^2 +...+2^2 - 1^2) - (10 - 9 +... + 2 - 1)=?$

We can substitute $10^2 - 9^2 +...+ 2^2 - 1^2$ with $5(6^2 - 5^2)$ which equals $5(11)$

We can also substitute $10 - 9 +...+ 2 - 1$ with $5(6 - 5)$ which equals $5(1)$

Our equation now is $5(11) - 5(1) = 5(10) = 50 =?$