Rom

reposted for readability

\(\text{Square A and Square B are both $2009$ by $2009$ squares.$\\$ Square A has both its length and width increased by an amount $x,\\$ while Square B has its length and width decreased by the same amount $x.\\~\\$ What is the minimum value of $x$ such that the difference in area between$\\$ the two new squares is at least as great as the area of a $2009$ by $2009$ square? }\)

\(\text{This is straightforward application of the binomial distribution}\\ p[\text{k answers correct}] = \dbinom{10}{k} \left(\dfrac 1 5\right)^k \left(\dfrac 4 5\right)^{10-k}\\~\\ p[0] = \left(\dfrac 4 5\right)^{10}\\ p[0\leq k \leq 3] = \sum \limits_{k=0}^3 p[k]\\ p[2\leq k \leq 10] = 1 - p[0\leq k \leq 1]= 1- p[0]-p[1]\\~\\ \text{I leave the plug and chug to you}\)

450 is the correct answer

f(4) = f(-4)

g(f(4)) = g(f(-4)) = 9

f(x) is an even function.

f(4) = f(-4)

so what's the answer?

\(\text{so we want to minimize $\left||x|-(-x^2-3x-2)\right|= \left||x| + x^2+3x+2\right|$}\)

\(\text{for $x<0$}\\ \left||x|+x^2+3x+2\right| = \left|x^2 +2x+2\right|= (x+1)^2+1\\ \text{The minimum of this length occurs at $x=-1$ and is length 1}\\~\\ \text{for $x\geq 0$}\\ \left||x|+x^2+3x+2\right| = \left|x^2 +4x+2\right| = \left|(x+2)^2-2\right|\\ \text{Since $x>0$ this has a minimum at $x=0$ and is length 2}\\ \text{Thus the smallest possible length is 1 which occurs at $x=-1$}\)

\(\text{we select two consecutive whole numbers and find their remainder after dividing by 4}\\ \text{the following pairs are possible}\\ (0,1),~(1,2),~(2,3)~(3,0)\\ \text{out of these 4 pairs of remainders 2 of them include 0, i.e.}\\ \text{the original pair had a number divisible by 4}\\ p = \dfrac 2 4 = \dfrac 1 2\)

\(\text{If you study this you find that you must have the pattern $\\MDDM$ embedded in the seating arrangement}\\ \text{There are 3 slots in which these 4 guests can be fit}\\ \text{Having fit them the 2 elves sit in the remaining seats}\\ \text{There are 4 ways to arrange $MDDM$}\\ \text{There are 2 ways to arrange the elves once the others are seated}\\ N = 3 \cdot 4 \cdot 2 = 24\)

answered on the thread at the link

\(\text{if $k>0$ and $x>1$ then $x^k>x$}\\ \text{if $k>0$ and $x=1$ then $x^k=1$}\\ \text{so the range of $f(x)$ for $x \in [1,\infty)$ is $f(x) \in [1,\infty)$}\)