Jim should receive 3/8 of the proceeds.

Slice the number XII = 12 across the middle to get VII = 7.

In base 2, 1 + 1 = 10

Where is "d"?

We have: \(y^2=x\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

           \(\Leftrightarrow y^2+1=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

           \(\Leftrightarrow y^2+1=\left(x^2+3x+1\right)^2\)

\(\Rightarrow y=0\) (Because \(y^2\) and \(\left(x^2+3x+1\right)^2\) is two consecutive square numbers)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\\x=-3\end{matrix}\right.\)

So x=0; x=-1; x=-2; x=-3.

When n has three digits, we have:

      \(n+S\left(n\right)\le999+27=1026\) (!)

\(\Rightarrow\) n has four digits.

Now we have: 

       \(\left(1000a+100b+10c+d\right)+\left(a+b+c+d\right)=1982\)

  \(\Leftrightarrow1001a+101b+11c+2d=1982\) (1)

\(\Rightarrow a=\left[\dfrac{1982}{1001}\right]=1\)

\(\left(1\right)\Leftrightarrow101b+11c+2d=981\) (2)

\(\Rightarrow b=\left[\dfrac{981}{101}\right]=9\)

 \(\left(2\right)\Leftrightarrow11c+2d=72\) (3)

\(\Rightarrow c=\left[\dfrac{72}{11}\right]=6\)

 \(\left(3\right)\Leftrightarrow2d=6\Leftrightarrow d=3\)

Because the factoring is only one 

So a=1; b=9; c=6; d=3.

Condition: \(x\ge82\)

We have: \(\sqrt{x+7}-\sqrt{x-82}=x-2017\)

          \(\Leftrightarrow\left(\sqrt{x+7}-45\right)+\left(44-\sqrt{x-82}\right)+2018-x=0\)

          \(\Leftrightarrow\dfrac{x-2018}{\sqrt{x+7}+45}+\dfrac{2018-x}{\sqrt{x-82}+44}+2018-x=0\)

          \(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{\sqrt{x-82}+44}-\dfrac{1}{\sqrt{x+7}+45}+1\right)=0\)

\(x\ge82\Rightarrow1>\dfrac{1}{\sqrt{x+7}+45}>\dfrac{1}{\sqrt{x-82}+44}>0\Rightarrow\dfrac{1}{\sqrt{x-82}+44}+1-\dfrac{1}{\sqrt{x+7}+45}>0\)

\(\Rightarrow2018-x=0\Rightarrow x=2018\)

So \(x=2018\).

abc is different 0 \(\Rightarrow\) a, b, c is different 0.

We have: \(\dfrac{x-b-c}{a}+\dfrac{x-a-c}{b}+\dfrac{x-a-b}{c}=3\)

          \(\Leftrightarrow\left(\dfrac{x-b-c}{a}-1\right)+\left(\dfrac{x-a-c}{b}-1\right)+\left(\dfrac{x-a-b}{c}-1\right)=0\)

          \(\Leftrightarrow\dfrac{x-a-b-c}{a}+\dfrac{x-a-b-c}{b}+\dfrac{x-a-b-c}{c}=0\)

          \(\Leftrightarrow\left(x-a-b-c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=0\)

          \(\Leftrightarrow x-a-b-c=0\) (because \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ne0\))

          \(\Leftrightarrow x=a+b+c\)

So \(x=a+b+c\)

We have: \(P=10a^2+10b^2+c^2\)

                    \(=\left(8a^2+\dfrac{1}{2}c^2\right)+\left(8b^2+\dfrac{1}{2}c^2\right)+\left(2a^2+2b^2\right)\)

                    \(\ge2\sqrt{8a^2.\dfrac{1}{2}c^2}+2\sqrt{8b^2.\dfrac{1}{2}c^2}+4\sqrt{a^2b^2}\) (AM-GM inequality)

                    \(=4ac+4bc+4ab\)

                    \(=4\left(ab+bc+ca\right)=4\)

So the minimum of P is 4 when \(a=b=\dfrac{1}{3};c=\dfrac{4}{3}\).

\(1000000!:999999!=\left(1.2.3.....1000000\right):\left(1.2.3.....999999\right)=1000000\)

We have: \(x^2\) is a square number

\(\Rightarrow x\in Z\)

-C₁: \(x⋮3\)

\(\Rightarrow x^2\equiv0\) (mod 3)

\(\Rightarrow x^2+5\equiv2\) (mod 3) (Impossible because \(x^2+5\) is a square number)

-C₂: \(x⋮̸3\)

\(\Rightarrow x^2\equiv1\) (mod 3)

\(\Rightarrow x^2-5\equiv2\) (mod 3) (Impossible because \(x^2-5\) is a square number)

 So can't find x.

\(\sqrt{2.3.4.5.6.7.10}=\sqrt{50400}=60\sqrt{14}\)

\(5-5.5+5:5=5-25+1=-19\)

1: \(12.37+12.7+12.6\)

  \(=444+84+72\)

  \(=600\)

2: \(12.37+12.7+12.6\)

 \(=12\left(37+7+6\right)\)

 \(=12.50\)

 \(=600\).

We have: \(y^2=\left(y+2018\right)^2\)

         \(\Leftrightarrow\left(y+2018\right)^2-y^2=0\)

         \(\Leftrightarrow2018\left(2y+2018\right)=0\)

         \(\Leftrightarrow2y=-2018\)

         \(\Leftrightarrow y=-1009\)

So \(y=-1009\).

We have: \(A=2018^2-2017^2+2016^2-2015^2+...+2^2-1^2\)  

                    \(=\left(2018-2017\right)\left(2018+2017\right)+\left(2016-2015\right)\left(2016+2015\right)+...+\left(2-1\right)\left(2+1\right)\)

                    \(=2018+2017+2016+2015+...+2+1\)

                    \(=\dfrac{2018.2019}{2}=2037171\).

So the last two digits of A is 71.