Problems

Is \(100! = 100\times 99 \times ...\times 2\) divisible by \(2^{100}\)?

Prove the magic trick for the number \(1089 = 33^2\): if you take any \(3\)-digit number \(\overline{abc}\) with digits coming in strictly descending order and subtract from it the number obtained by reversing the digits of the original number \(\overline{abc} - \overline{cba}\) you get another \(3\)-digit number, call it \(\overline{xyz}\). Then, no matter which number you started with, the sum \(\overline{xyz} + \overline{zyx} = 1089\).
Recall that a number \(\overline{abc}\) is divisible by \(11\) if and only if \(a-b+c\) also is.

Do there exist two numbers such that their sum, quotient and product would be all equal to each other?

It is easy to construct one equilateral triangle using three identical matches. Is it possible to construct four equilateral triangles by adding just three more matches identical to the original ones?

Find the largest possible number of bishops that can be placed on the \(8 \times 8\) chessboard so that no two bishops threaten each other.

There are \(24\) children in the class and some of them are friends with each other. The following rules apply:

• If someone (say Alice) is a friend with someone else (say Bob), then the second student (Bob) is also a friend with the first (Alice).

• If Alice is friend with Bob and Bob is friend with Claire, then Alice is also friend with Claire.

Find a misconception in the following statement: under the above conditions Alice is friend with herself.

Theorem: All people have the same eye color.

"Proof" by induction: This is clearly true for one person.

Now, assume we have a finite set of people, denote them as \(a_1,\, a_2,\, ...,\,a_n\), and the inductive hypothesis is true for all smaller sets. Then if we leave aside the person \(a_1\), everyone else \(a_2,\, a_3,\,...,\,a_n\) has the same color of eyes and if we leave aside \(a_n\), then all \(a_1,\, a_2,\,a_3,...,\,a_{n-1}\) also have the same color of eyes. Thus any \(n\) people have the same color of eyes.
Find a mistake in this "proof".

Let’s prove that \(1\) is the largest natural number.
Let \(n\) be the largest natural number. Then, \(n^2\), being a natural number, is less than or equal to \(n\). Therefore \(n^2-n=n(n-1)\leq 0\). Hence, \(0\leq n\leq 1\). Therefore \(n=1\).

Recall that \((n+1)^2=n^2+2n+1\) and after expansion we get \((n+1)^2-(2n+1)=n^2\). Subtract \(n(2n+1)\) from both sides \((n+1)^2-(2n+1)-n(2n+1)=n^2-n(2n+1)\) and rewrite it as \((n+1)^2-(n+1)(2n+1)=n^2-n(2n+1)\).
Now we add \(\frac{(2n+1)^2}{4}\) to both sides: \((n+1)^2-(n+1)(2n+1)+\frac{(2n+1)^2}{4}=n^2-n(2n+1)+\frac{(2n+1)^2}{4}\).
Factor both sides into square: \(((n+1)-\frac{2n+1}{2})^2=(n-\frac{2n+1}{2})^2\).
Now take the square root: \((n+1)-\frac{2n+1}{2}=n-\frac{2n+1}{2}\).
Add \(\frac{2n+1}{2}\) to both sides and we get \(n+1=n\) which is equivalent to \(1=0\).

This problem is often called "The infinite chocolate bar". Depicted below is a way to get one more piece of chocolate from the \(5\times 6\) chocolate bar. Do you see where is it wrong?