BD-BDMO 2016

(a) Show that $n(n + 1)(n + 2)$ is divisible by $6$ .

(b) Show that $1^{2015} + 2^{2015} + 3^{2015} + 4^{2015} + 5^{2015} + 6^{2015}$ is divisible by $7$ .

(a) How many positive integer factors does $6000$ have?

(b) How many positive integer factors of $6000$ are not perfect squares?

Consider the set of integers $\left \{ 1, 2, \dots , 100 \right \}$ . Let $\left \{ x_1, x_2, \dots , x_{100} \right \}$ be some arbitrary arrangement of the integers $\left \{ 1, 2, \dots , 100 \right \}$ , where all of the $x_i$ are different. Find the smallest possible value of the sum $$S = \left | x_2 - x_1 \right | + \left | x_3 - x_2 \right | + \cdots+ \left |x_{100} - x_{99} \right | + \left |x_1 - x_{100} \right | .$$

Suppose there are $m$ Martians and $n$ Earthlings at an intergalactic peace conference. To ensure the Martians stay peaceful at the conference, we must make sure that no two Martians sit together, such that between any two Martians there is always at least one Earthling.

(a) Suppose all $m + n$ Martians and Earthlings are seated in a line. How many ways can the Earthlings and Martians be seated in a line?

(b) Suppose now that the $m+n$ Martians and Earthlings are seated around a circular round-table. How many ways can the Earthlings and Martians be seated around the round-table?

$\triangle ABC$ is an isosceles triangle with $AC = BC$ and $\angle ACB < 60^{\circ}$ . $I$ and $O$ are the incenter and circumcenter of $\triangle ABC$ . The circumcircle of $\triangle BIO$ intersects $BC$ at $D \neq B$ .

(a) Do the lines $AC$ and $DI$ intersect? Give a proof.

(b) What is the angle of intersection between the lines $OD$ and $IB$ ?

Juli is a mathematician and devised an algorithm to find a husband. The strategy is: • Start interviewing a maximum of $1000$ prospective husbands. Assign a ranking $r$ to each person that is a positive integer. No two prospects will have same the rank $r$ . • Reject the first $k$ men and let $H$ be highest rank of these $k$ men. • After rejecting the first $k$ men, select the next prospect with a rank greater than $H$ and then stop the search immediately. If no candidate is selected after $999$ interviews, the $1000th$ person is selected.

Juli wants to find the value of $k$ for which she has the highest probability of choosing the highest ranking prospect among all $1000$ candidates without having to interview all $1000$ prospects.(a) (6 points:) What is the probability that the highest ranking prospect among all $1000$ prospects is the $(m + 1)th$ prospect?(b) (6 points:) Assume the highest ranking prospect is the $(m + 1)th$ person to be interviewed. What is the probability that the highest rank candidate among the first $m$ candidates is one of the first $k$ candidates who were rejected?(c) (6 points:) What is the probability that the prospect with the highest rank is the $(m+1)th$ person and that Juli will choose the $(m+1)th$ man using this algorithm?(d) (16 points:) The total probability that Juli will choose the highest ranking prospect among the $1000$ prospects is the sum of the probability for each possible value of $m+1$ with $m+1$ ranging between $k+1$ and $1000$ . Find the sum. To simplify your answer use the formula $In N \approx \frac{1}{N-1}+\frac{1}{N-2}+...+\frac{1}{2}+1$ (e) (6 points:) Find that value of $k$ that maximizes the probability of choosing the highest ranking prospect without interviewing all $1000$ candidates. You may need to know that the maximum of the function $x ln \frac{A}{x-1}$ is approximately $\frac{A + 1}{e}$ , where $A$ is a constant and $e$ is Euler's number, $e = 2.718....$

Consider the integral $Z(0)=\int^{\infty}_{-\infty} dx e^{-x^2}= \sqrt{\pi}$ . (a) Show that the integral $Z(j)=\int^{\infty}_{-\infty} dx e^{-x^{2}+jx}$ , where $j$ is not a function of $x$ , is $Z(j)=e^{j^{2}/4a} Z(0)$ .(b) Show that $$\dfrac 1 {Z(0)}=\int x^{2n} e^{-x^2}= \dfrac {(2n-1)!!}{2^n},$$ where $(2n-1)!!$ is defined as $(2n-1)(2n-3)\times\cdots\times3\times 1$ .(c) What is the number of ways to form $n$ pairs from $2n$ distinct objects? Interpret the previous part of the problem in term of this answer.