191

Total Ways = $$$\binom{14}{5}$$$. Out of $$$14$$$ places choose $$$5$$$ places for men. Women will follow.
There are total $$$7$$$ pairs of seats. Choose $$$5$$$ out of them. For each pair of seats there are $$$2$$$ ways.
So required ways = $$$\binom{7}{5} \cdot 2^5$$$
Required Probability = $$$\frac{\binom{7}{5} \cdot 2^5}{\binom{14}{5}}= \frac{48}{143}$$$.

607

Two lines can only intersect once.
Maximum number of intersections = $$$\binom{n}{2}$$$ (Choose any two lines and they intersect)
Maximum number of intersections occur when for each intersection point there are only two lines intersecting at that point. Lets label these $$$T2$$$ intersection points.
If there are intersection points where there are $$$x$$$ ($$$x$$$ > $$$2$$$) lines intersecting at some point, then we will lose $$$T2$$$ points.
Amount of point we lost = Amount of points $$$x$$$ lines could have contributed = $$$\binom{x}{2}$$$
So the final answer becomes
$$$\binom{40}{2} - 3\binom{3}{2} - 4\binom{4}{2}- 5\binom{5}{2}- 6\binom{6}{2} = \boxed{607}.$$$

12

Power of $$$x$$$ in $$$n!$$$ = $$$\lfloor \frac{n}{x} \rfloor + \lfloor \frac{n}{x^{2}} \rfloor+ \lfloor \frac{n}{x^3{}} \rfloor + ...$$$ Using this formula we get.. $$$13! = 2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
For a number to be perfect square it prime factorisation should consist of even powers.
Let's write it even and odd powers seperately $$$13! = 2^{10} \cdot 3^{4} \cdot 5^{2} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
Now we want sum of all numbers $$$2^{x} \cdot 3^{y} \cdot 5^{z} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
such that $$$0 \leq x \leq 10 $$$ and $$$0 \leq y \leq 4$$$ and $$$0 \leq z \leq 2$$$ and $$$x, y, z$$$ are even
The resultant number will be $$$\sum\limits_{x} \sum\limits_{y} \sum\limits_{z}2^{x} \cdot 3^{y} \cdot 5^{z} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
which is
$$$(\sum\limits_{x} (2^{x}))\cdot (\sum\limits_{y} (3^{y})) \cdot (\sum\limits_{z} (5^{z})) \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
This can be computed using GP series.
final answer = $$$ 2^{1} \cdot 3^{2} \cdot 5^{1} \cdot 7^{3} \cdot 11^{1} \cdot 13^{4}$$$

944

The key idea is that the summation without the floors should be close to $$$0$$$ because each floor can only change the sum by less than $$$1$$$. The natural choice is a value of $$$a$$$ near the one that makes the aforementioned sum exactly $$$0$$$. Luckily, we compute [ \sum_{n=1}^{2023}\frac{n^2-na}5 = \frac15 \left( \frac{2023 \cdot 2024 \cdot 4047}{6} — a \frac{2023 \cdot 2024}{2} \right), ]which is zero at $$$a = \frac{4047}{3} = 1349$$$, an integer. Trying $$$a = 1349$$$ for the value of $$$U$$$, we have [ \begin{aligned} U=\sum_{n=1}^{2023}\left\lfloor\frac{n^2-1349n}5\right\rfloor &= \sum_{n=1}^{2023}\left(\frac{n^2-1349n}5 — \frac{(n^2 — 1349n) \mod 5}5 \right) \ &= -\frac15 \sum_{n=1}^{2023} (n^2 — 1349n) \mod 5 \end{aligned}]But note that $$$n^2 - 1349n \equiv n^2 + n \pmod 5$$$ takes on the following values at the residues mod $$$5$$$, \begin{tabular}{c|c} $$$n$$$ & $$$(n^2 + n) \mod 5$$$ \ \hline $$$1$$$ & $$$2$$$ \ $$$2$$$ & $$$1$$$ \ $$$3$$$ & $$$2$$$ \ $$$4$$$ & $$$0$$$ \ $$$0$$$ & $$$0$$$ \end{tabular} Finally, note that $$$n=1$$$ to $$$n=2023$$$ passes through almost $$$405$$$ periods of length $$$5$$$, only missing two zeroes that don't matter, with each period summing to $$$5$$$, so [ U = -\frac15 (405 \cdot 5) = -405.]It follows that $$$a + U = 1349 - 405 = \boxed{944}$$$.

Math problems that aim to aid competitive programming skills.
Target audience : Experts and below.