Is it possible to colour the cells of a \(3\times 3\) board red and yellow such that there are the same number of red cells and yellow cells?
A group of \(15\) elves decided to pay a visit to their relatives in a distant village. They have a horse carriage that fits only \(5\) elves. In how many ways can they assemble the ambassador team, if at least one person in the team needs to be able to operate the carriage, and only \(5\) elves in the group can do that?
There are \(5\) pirates and they want to share \(8\) identical gold coins. In how many ways can they do it if each pirate has to get at least one coin?
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.
We want to wrap \(12\) Christmas presents in different coloured paper. We have \(6\) different patterns of paper and we want to use each one exactly twice. In how many ways can we do this?
Mr Roberts wants to place his little stone sculptures of vegetables on the different shelves around the house. He has \(17\) sculptures in total and three shelves that can fit \(7\), \(8\) and \(2\) sculptures respectively. In how many ways can he do this?
The order of sculptures on the shelf does not matter.
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 \(1=0\).