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4

Your argument is incorrect. Suppose that each boy plucked a different number of apples. Then the smallest total the group of boys could have plucked is $$0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 = 105$$ However, they collectively plucked just $100$ apples. Hence, two boys must have plucked the same number of apples.


4

Let the colors used be $A$, $B$, $C$, $D$. We call an unordered pair of squares neoliberal if the two squares lie in the same row and are of different colors. Every row gives rise to $6 \cdot 25^2$ neoliberal pairs (given by $\binom{4}{2}$ possible pairs of colors and $25$ squares of each color). Thus, summing over all the rows, there is a total of $100 ...


3

You already proved the existence of a column-pair such that $76$ of its rows contains different colored square pairs. Four different colored squares that you're looking for is actually in this column-pair: Number the colors as $0$, $1$, $2$ and $3$. Aforementioned $76$ rows can only have six different pairs: $A=\{0,1\}$, $A'=\{2,3\}$, $B=\{0,2\}$, ...


3

Here's an outline of the proof: Show that there are $14$ primes under $\sqrt{2015}$. Let $R$ be a set of $14$ composite coprime numbers under $2015$. It is well-known that any composite number has a prime factor under its square root, so in this case, each of these composite numbers has one of the prime factors under $\sqrt {2015}$. Now, using that ...


3

Divide into 25 smaller squares, each side $\frac{1}{5}$. At least one square must contain at least 3 points. The diagonal of a small square is $\frac{\sqrt{2}}{5}=\sqrt{0.08}$, so each of the 3 points is a distance less than that from each of the other 2.


3

Perhaps I will write something wrong? (This seems easy to me). First of all, I think you mean there can't exist an injective function $f:I_m\to I_n$, for if $n<m$ there actually exists an injective $I_n\to I_m$ (the inclusion). Suppose $n<m$ and $f:I_m\to I_n$ injective. Then $f:I_m\to f(I_m)$ is bijective, where $f(I_m)=\{f(1),...,f(m)\}$. Clearly ...


2

Hint: The base case is easy, so let's look at the inductive step. Assuming $P(m)$ is true, let's consider any set $S$ of $2m+3$ distinct integers from among $[-2m-1, 2m+1]$. If less than $3$ of those are from among $R = \{\pm(2m+1), \pm2m\}$, the induction hypothesis takes care of it. So that leaves cases where $|S \cap R| \in \{3, 4\}$. Here use ...


1

Some of the better "pigeon hole" problems are quite subtle and it is not trivial to figure out how the principle will be used. In this case, think of all the different sums you can create. (How many subsets, of all possible cardinalities, does a set with 10 elements have?) Those are the pigeons. On the other hand, the sums are at most 1,000 (ten numbers, ...



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