Problems

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.

Does there exist a power of \(3\) that ends in \(0001\)?

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 \(n=1\).

Theorem: If we mark \(n\) points on a circle and connect each point to every other point by a straight line, the lines divide the interior of the circle is into is \(2n-1\) regions.
"Proof": First, let’s have a look at the smallest natural numbers.

• When \(n=1\) there is one region (the whole disc).

• When \(n=2\) there are two regions (two half-discs).

• When \(n=3\) there are \(4\) regions (three lune-like regions and one triangle in the middle).

• When \(n=4\) there are \(8\) regions, and if you’re still not convinced then try \(n=5\) and you’ll find \(16\) regions if you count carefully.

Our proof in general will be by induction on \(n\). Assuming the theorem is true for \(n\) points, consider a circle with \(n+1\) points on it. Connecting \(n\) of them together in pairs produces \(2n-1\) regions in the disc, and then connecting the remaining point to all the others will divide the previous regions into two parts, thereby giving us \(2\times (2n-1)=2n\) regions.

Let’s "prove" that the number \(1\) is a multiple of \(3\). We will use the symbol \(\equiv\) to denote "congruent modulo \(3\)". Thus, what we need to prove is that \(1\equiv 0\) modulo \(3\). Let’s see: \(1\equiv 4\) modulo \(3\) means that \(2^1\equiv 2^4\) modulo \(3\), thus \(2\equiv 16\) modulo \(3\), however \(16\) gives the remainder \(1\) after division by \(3\), thus we get \(2\equiv 1\) modulo \(3\), next \(2-1\equiv 1-1\) modulo \(3\), and thus \(1\equiv 0\) modulo \(3\). Which means that \(1\) is divisible by \(3\).

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 \(1=0\).

Look at the following diagram, depicting how to get an extra cell by reshaping triangle.

Can you find a mistake? Certainly the triangles have different area, so we cannot obtain one from the other one by reshaping.

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?