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5

In this diagram, the large triangle has sides of $2$ while the four smaller ones have sides of $1$. So place five pandas into those four small triangles. The maximum distance between any two points in any small triangle is clearly just $1$. Use the pigeonhole principle and finish from here.


3

This is a really cool variant on the Monte Hall "do you choose the other door" problem, or for bridge enthusiasts, the principle of restricted choice. Let me rephrase the issue as follows: $A$ is presented with three independent fair-coin randoms (behind three "doors"). The rules of the game are that $A$ must present $B$ with two of those randoms by ...


3

Let's count, chosen one column, how many different pairs of points we can have in that column: this is $ {13 \choose 2}=78.$ Call $a_i$ the number of points that appear in one column, we have that $$a_1+a_2+...+a_{13}=53.$$ In the column $i$ there are $a_i \choose 2$ different column-pairs of points, if we prove that $${a_1 \choose 2}+{a_2 \choose ...


3

There are $50$ pairs of diametrically opposite seats. You have more than $50$ women.... Does this make you think of stuffing birds in holes?


3

Ross's argument shows that to achieve four you'd have to have each book bought by exactly four people. That isn't possible, since there are exactly 30 purchases, and 30 isn't divisible by $4$. Reproducing Ross's logic here, with my edits: Given a person who bought books $a,b,c$, there are nine other people, so either each of the books $a,b,c$ were ...


2

At a round table with 100 seats, there are 50 pairs of seats diametrically opposite to each other in the pair. The number of women $n$ to distribute among the pairs is greater than the number of pairs, $m$; i.e., $n > m$. By the pigeonhole principle, at least one pair contains more than one woman.


2

Hint: Let $a_k$ be the number of marbles the person has been given after $k$ days. Since the person receives at least one marble each day, each element in the set $$A = \{a_1, a_2, \ldots, a_{49}\}$$ is distinct. Since the person receives at most eleven marbles in a week and receives marbles for seven weeks, $$A \subseteq \{1, 2, \ldots, 77\}$$ Let ...


2

However, suppose that I claim that the probability should be 1/2, and here is my reasoning: First, I point out that there are three coin tosses and only two possible outcomes for each toss. Therefore, at least two out of the three tosses are guaranteed to have the same result. I need only point out that the probability of the third toss having the same ...


2

Don't be caught up in the definition - think of it using basic logic. Come up with a worst-case scenario. We could draw the $10$ red pencils, followed by the $8$ blue pencils, followed by the $8$ green pencils, and without ever drawing a yellow pencil, we have already used $10 + 8 + 8 = 26$ moves. But we still need one more draw to get all the colors. Our ...


2

Given any $40$ people, at least four of them were born in the same month of the year. The the phrase "at least four of them were born in the same month of the year" means just that there is some subset of four people out of the $40$ who share a common birth month. It does not mean that everyone belongs to some clique of four people with the same birth ...


1

number of boxes I believe should be 27, as you need to choose at least 27 pencils to guarantee one of each, lets look at it like this in the worst case you have after 26 choices, 10 reds, 8 blue, 8 green for 26, this you need to draw 1 more ball with a 100% chance of getting yellow (as no others are left), any other combination less than 27 then would not ...


1

Hint: Suppose otherwise. Then at most three people are born in each of the twelve months of the year. What can you conclude?


1

Hint: you can cut the pen into four triangles with side $1$


1

The theorem is false. It becomes true if you change "XOR" to "OR" and your proof becomes correct if you replace $\Leftarrow\Rightarrow$ with $\wedge$.


1

For the $i$-th $1$ in the matrix, define $x_i$ to be the number of $1$'s to the left of that $1$. So $\sum_{i=1}^{53} x_i$ is the number of $1 \times 2$ all-$1$ submatrices. If this number is more than $\binom{13}{2}=78$, we have a pair of columns containing two $1 \times 2$ all-$1$ submatrices, i.e., a rectangle. Since we have $13$ rows, for any $k \in ...


1

First of all, let us consider that no two such women exist, as mentioned in the question. Say, we consider any $2$ diametrically opposite sitting people as an ordered pair. So we will have $50$ such ordered pairs as for example, $(P_1,P_{51}),(P_2,P_{52}),(P_3,P_{53}),(P_4,P_{54}), \ldots (P_{50},P_{100})$. If every one persons from those $50$ pairs have ...



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