Rom

\(\text{I'm going to assume you can't use calculus}\\ p=200 = 2 \ell + 2w\\ w = \dfrac{200-2\ell}{2} \\ area = \ell \cdot w =\\ \ell \left(\dfrac{200-2\ell}{2}\right) = \\ 100\ell - \ell^2 = \\ -(\ell^2 - 100\ell + 2500 - 2500) = \\ -(\ell -50)^2 + 2500 \\ \text{and this has it's maximum value, }2500 \text{ at }\ell=50,~w=50 \\ \)

\(\text{can you use calculus?}\\ \text{If so we can find the first derivative of the height (it's vertical velocity)}\\ \text{and solve the the }t \text{ that makes this zero}\\ \text{(the ball stops for a moment at the top of it's arc)}\\ v_z(t) = \dfrac{d}{dt} h(t) = -32t + 64\\ v_z(t) = 0 \Rightarrow t = 2\\ h(2) = -16(4) + 64(2) + 31 = 95\)

\(\text{If you can't use calculus (and I suspect you can't)}\\ \text{we can rewrite }h(t) \text{ in standard parabola form}\\ -16t^2 + 64t + 31 = \\ -16(t^2 -4t) + 31 = \\ -16((t-2)^2-4)+31 = \\ -16(t-2)^2 + 64 + 31 = \\ -16(t-2)^2 + 95\\ \text{and it should be clear this reaches a maximum of }\\ h=95 \text { at }t=2\)

\(3x^2 + b x + 10 = \\ 3\left(x^2 + \dfrac b 3 x + \dfrac{10}{3}\right) = \\ 3\left(\left(x+\dfrac b 6\right)^2 - \dfrac{b^2}{36}+\dfrac{10}{3}\right) = \\ 3\left(x+\dfrac b 6\right)^2+\left(10-\dfrac{b^2}{12}\right)\)

\(m = \dfrac b 6\\ \text{so suppose that 6 is the largest integer that must be a divisor of }b\\ \dfrac{b^2}{36} = \dfrac{(6k)^2}{36} = \dfrac{36k^2}{36}=k^2 \in \mathbb{Z}\\ \text{so }n \in \mathbb{Z}\\ \text{So the number you are after is }6\)

\(\text{as }a \text{ and }b \text{ are solutions to }x^2 + ax + b = 0\\ \text{we can write}\\ x^2 + ax + b = (x-a)(x-b) = x^2 -(a+b)x + ab\\ a = -(a+b) \text{ and } b= a b\\ b = -2a\\ b = a(-2a) = -2a^2\\ -2a = -2a^2 \Rightarrow a = 1\\ b = -2a = -2\\ (a,b) = (1,-2) \)

each step can be denoted as (1,0,0), (0,1,0), (0,0,1)

the sum of the steps equals (2,2,2)

you are not allowed to overshoot in any direction

It should be pretty clear we need to take 2 steps in each direction.

So the total number of paths is just

\(N=\dbinom{6}{2}\dbinom{4}{2} = 90\)

The first term represents choosing two easts say, from the 6 possible steps.

The second term represents choosing two norths from the remaining 4 steps.

Give those the two up steps are fixed.

It doesn't matter which direction you pick for choosing first and second.  The results are the same.

As far as I can see the maximum mean score you can achieve with these conditions is 14.

\(n=2^k,~k=0,1,\dots,6\)

\(\text{break the figure into 3 pieces}\\ \text{the first piece is 2x1001 white squares to the left of the shaded squares}\\ \text{the second piece is the 2 shaded squares}\\ \text{the third is the 2x1001 white squares to the right of the shaded squares}\\ \text{to pick a rectangle that does not contain the shaded squares we pick two points}\\ \text{from either the first section or the third section and these define our rectangle}\\ \text{there are }\dbinom{1001}{2} =500500\text{ ways to pick 2 points in both sections}\\ \text{so there are }2 \cdot 500500 = 1001000 \text{ ways to pick the rectangles that contain no shaded squares}\)

\(\text{There are }\dbinom{2003}{2} = 2005003 \text{ total ways to pick 2 points from the 2003}\\ \text{thus }\\ P[\text{rectangle contains no shaded square}]= \dfrac{1001000}{2005003 } = \dfrac{1000}{2003}\approx 0.4992\)

\(\dfrac{\sum \limits_{k=1}^{1005}~k x^{2k}}{\sum \limits_{j=1}^{1005}~2jx^{2j-1}}=\\ \dfrac{x}{2}\dfrac{\sum \limits_{k=1}^{1005}~2k x^{2k}}{\sum \limits_{j=1}^{1005}~2jx^{2j}}=\\ \dfrac x 2\)

\(5(2a-5+b)\)