Carelessly Cake

Denote $A=x^2+1$ , $B=x^2-1$ , $C=2x$ , $D=x^2+2x$ , $E=3x-5$ .

Suppose the missing number (or the not chosen number) is $S_M$ , the extra number (the number that was added twice) is $S_X$ .

Then, $A+B+C+D+E=3x^2+7x-5=47+S_M-S_X$ , $S=47-S_X$ .

Define $n=S_M-S_X$ , then $n=3x^2+7x-52$ , thus we know that $n$ must be even.

Now study the table below: $\begin{array}{|c|c|c|} \hline \\ & \text{When } x \text{ is odd} & \text{When } x \text{ is even}\\ \hline A & \text{even} & \text{odd} \\ \hline B & \text{even} & \text{odd} \\ \hline C & \text{even} & \text{even} \\ \hline D & \text{odd} & \text{even} \\ \hline E & \text{even} & \text{odd} \\ \hline \end{array}$

Pretty cool.

Consider all the possible values of $n$ , there are only 14 possible values for $n$ : $\pm (A-B),\quad\pm (A-C),\quad\pm (A-E),\quad\pm (B-C),$ $\pm (B-E),\quad\pm (C-D),\quad\pm (C-E)$ [ $n$ cannot be $\pm (A-D)$ , $\pm (B-D)$ and $\pm (D-E)$ because that will definitely make $n$ odd]

Let $n=ax^2+bx+c$ , then $3x^2+7x-52-n=(3-a)x^2+(7-b)x-(52+c)=0$ , consider all the possible ordered triples $(a,b,c)$ : $\begin{array}{|c|c|c|c|c|c|} \hline \\ n & a & b & c & a+b+c & a-b+c\\ \hline A-B & 0 & 0 & 2 & 2 & 2 \\B-A & 0 & 0 &-2 & -2 & -2\\ \hline A-C & 1 & -2 & 1 & 0 & 4\\C-A & -1 & 2 & -1 & 0 & -4\\ \hline A-E & 1 & -3 & 6 & 4 & 10\\E-A & -1 & 3 & -6 & -4 & -10\\ \hline B-C & 1 & -2 & -1 & -2 & 2\\C-B & -1 & 2 & 1 & 2 & -2\\ \hline B-E & 1 & -3 & 4 & 2 & 8 \\E-B & -1 & 3 & -4 & -2 & -8\\ \hline C-D & -1 & 0 & 0 & -1 & -1\\D-C & 1 & 0 & 0 & 1 & 1\\ \hline C-E & 0 & -1 & 5 & 4 & 6\\E-C & 0 & 1 & -5 & -4 & -6\\ \hline \end{array}$ Wow!

See that $-1\leqslant a\leqslant1$ , $-3\leqslant b\leqslant3$ , $-6\leqslant c\leqslant6$ .

From $(3-a)x^2+(7-b)x-(52+c)=0$ , we get $-6\leqslant c=(3-a)x^2+(7-b)x-52\leqslant6 \\ 46\leqslant (3-a)x^2+(7-b)x\leqslant58$

Its easy to figure out that $x\neq0$ , now let's split it into 2 cases:

Case 1: $x>0$

If $x>0$ , then $46\leqslant (3-a)x^2+(7-b)x<4x^2+10x \\ x(2x+5)>23 \\ \therefore x\geqslant3$ $58\geqslant (3-a)x^2+(7-b)x>2x^2+4x \\ x(x+2)<29 \\ \therefore x\leqslant4$ $\therefore 3\leqslant x\leqslant4$

When $x=3$ , $\begin{aligned}-c&=9(a-3)+3(b-7)+52\\&=9a+3b+4\\&\equiv a-b\pmod{4}\end{aligned} \\ \therefore a-b+c\equiv0\pmod{4}\qquad(1)$ Meanwhile, $c\equiv-1\pmod{3}\qquad(2)$ As we can see from the table above, when $x=3$ , $n$ satisfies $(1)$ and $(2)$ if $n=C-A$ or $n=E-B$ . In either case, they both satisfy $-c=9a+3b+4$ .

$\therefore n=C-A,\,E-B$ $\begin{aligned}\therefore S_X&=A,B\\&=x^2+1,\,x^2-1\\&=10,\,8\end{aligned}$ $\begin{aligned}S&=47-S_X\\&=37,\,39\end{aligned}$

When $x=4$ , $\begin{aligned}-c&=16(a-3)+4(b-7)+52\\&=16a+4b-24\\&\equiv a+b\pmod{3}\end{aligned} \\ \therefore a+b+c\equiv0\pmod{3}\qquad(3)$ Meanwhile, $c\equiv0\pmod{4}\qquad(4)$ However, when $x=4$ , from the table above, there's no such $n$ that satisfy both $(3)$ and $(4)$ .

Case 2: $x<0$

If $x<0$ , then $46\leqslant (3-a)x^2+(7-b)x<4x^2+4x \\ 2x(x+1)\geqslant23 \\ \therefore x\leqslant -4$ $58\geqslant (3-a)x^2+(7-b)x\geqslant2x^2+10x \\ x(x+5)\leqslant29 \\ \therefore x\geqslant -8$ $\therefore -8\leqslant x\leqslant-4$

When $x=-8$ , $\begin{aligned}-c&=64(a-3)-8(b-7)+52\\&\equiv a-3-b+7+3 \\&\equiv a-b\pmod{7}\end{aligned} \\ \therefore a-b+c\equiv0\pmod{7}\qquad(5)$ Obviously no such $n$ satisfies $(5)$ when $x=-8$ .

When $x=-7$ , $\begin{aligned}-c&=49(a-3)-7(b-7)+52\\&\equiv a-3+b-7+4 \\&\equiv a+b+2\pmod{8}\end{aligned} \\ \therefore a+b+c\equiv6\pmod{8}\qquad(6)$ $c\equiv3\pmod{7}\qquad(7)$ No such $n$ satisfies both $(6)$ and $(7)$ when $x=-7$ .

When $x=-6$ , $\begin{aligned}-c&=36(a-3)-6(b-7)+52\\&\equiv a-3+b-7+3 \\&\equiv a+b\pmod{7}\end{aligned} \\ \therefore a+b+c\equiv6\pmod{8}\qquad(8)$ $c\equiv3\pmod{7}\qquad(9)$ No such $n$ satisfies both $(8)$ and $(9)$ when $x=-6$ .

When $x=-5$ , $\begin{aligned}-c&=25(a-3)-5(b-7)+52\\&\equiv a-3-b+7 \\&\equiv a-b\pmod{4}\end{aligned} \\ \therefore a-b+c\equiv0\pmod{4}\qquad(10)$ $c\equiv2\pmod{5}\qquad(11)$ No such $n$ satisfies both $(10)$ and $(11)$ when $x=-5$ .

When $x=-4$ , $\begin{aligned}-c&=16(a-3)-4(b-7)+52\\&=16a-4b+32\\&\equiv a-3-b+7 \\&\equiv a-b\pmod{4}\end{aligned} \\ \therefore a-b+c\equiv0\pmod{4}\qquad(12)$ $a+b+c\equiv3\pmod{5}\qquad(13)$ When $x=-4$ , if $n=E-B$ , $a=-1$ , $b=3$ , $c=-4$ satisfies $-c=16a-4b+32$ , $(12)$ and $(13)$ . $\therefore n=E-B$ $\begin{aligned}\therefore S_X&=B\\&=x^2-1\\&=15\end{aligned}$ $\begin{aligned}S&=47-S_X\\&=32\end{aligned}$

Therefore, the sum of all possible values of $S$ is $39+37+32=108$ .