Problems

Scrooge McDuck has \(100\) golden coins on his office table. He wants to distribute them into \(10\) piles so that no two piles contain the same amount of coins. Moreover, no matter how you divide any of the piles into two smaller piles, among the resulting \(11\) piles there will be two with the same amount of coins. Find an example of how he could do that.

In a certain state, there are three types of citizens:

• A fool considers everyone a fool and themselves smart;

• A modest clever person knows truth about everyone’s intellectual abilities and consider themselves a fool;

• A confident clever person knows about everyone intellectual abilities correctly and consider themselves smart.

There are \(200\) deputies in the High Government. The Prime Minister conducted an anonymous survey of High Government members, asking how many smart people are there in the High Government. After reading everyone’s response he could not find out the number of smart people. But then the only member who did not participate in the survey returned from the trip. They filled out a questionnaire about the entire Government including themselves and after reading it the Prime Minister understood everything. How many smart could there be in the High Government (including the traveller)?

Find the mistake in the sequence of equalities: \(-1=(-1)^{\frac{2}{2}}=((-1)^2)^{\frac{1}{2}}=1^{\frac{1}{2}}=1\).

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 \(3\).

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.

Consider the following "proof" that any triangle is equilateral: Given a triangle \(ABC\), we first prove that \(AB = AC\). First let’s draw the bisector of the angle \(\angle A\). Now draw the perpendicular bisector of segment \(BC\), denote by \(D\) the middle of \(BC\) and by \(O\) the intersection of these lines. See the diagram

Draw the lines \(OR\) perpendicular to \(AB\) and \(OQ\) perpendicular to \(AC\). Draw lines \(OB\) and \(OC\). Then the triangles, \(RAO\) and \(QAO\) are equal, since we have equal angles \(\angle ORA = \angle OQA = 90°,\) and \(\angle RAO = \angle QAO,\) and the common side \(AO\). On the other hand the triangles \(ROB\) and \(QOC\) are also equal since the angles \(\angle BRO = \angle CQO = 90°\), the hypotenuses \(BO = OC\) the legs \(RO = OQ\). Thus, \(AR = AQ,\) \(RB = QC,\) and \(AB = AR + RB = AQ + QC = AC.\) Q.E.D.

As a corollary, one can show that all the triangles are equilateral, by showing that \(AB = BC\) in the same way.

Let’s prove the following statement: every graph without isolated vertices is connected.
Proof We use the induction on the number of vertices. Clearly the statement is true for graphs with \(2\) vertices. Now, assume we have proven the statement for graphs with up to \(n\) vertices.
Take a graph with \(n\) vertices by induction hypothesis it must be connected. Let’s add a non-isolated vertex to it. As this vertex is not isolated, it is connected to one of the other \(n\) vertices. But then the whole graph of \(n+1\) vertices is connected!

Let’s compute the infinite sum: \[1+2 + 4 + 8 + 16 + ... + 2^n + ... = c\] Observe that \(1+2+4+8+... = 1 + 2(1+2+4+8+16+...)\), namely \(c = 1+2c\), then it follows that \[c = 1+2+4+8+... = -1.\]

Let’s prove that any \(90^{\circ}\) angle is equal to any angle larger than \(90^{\circ}\). On the diagram

We have the angle \(\angle ABC = 90^{\circ}\) and angle \(\angle BCD> 90^{\circ}\). We can choose a point \(D\) in such a way that the segments \(AB\) and \(CD\) are equal. Now find middles \(E\) and \(G\) of the segments \(BC\) and \(AD\) respectively and draw lines \(EF\) and \(FG\) perpendicular to \(BC\) and \(AD\).
Since \(EF\) is the middle perpendicular to \(BC\) the triangles \(BEF\) and \(CEF\) are equal which implies the equality of segments \(BF\) and \(CF\) and of angles \(\angle EBF = \angle ECF\), the same about the segments \(AF=FD\). By condition we have \(AB=CD\), thus the triangles \(ABF\) and \(CDF\) are equal, thus \(\angle ABF = \angle DCF\). But then we have \[\angle ABE = \angle ABF + \angle FBE = \angle DCF + \angle FCE = \angle DCE.\]