There are 13 green, 15 blue, and 17 red chameleons on an island. Whenever two chameleons of different colours meet, they both swap to the third colour (i.e., a green and blue would both become red). Is it possible for all chameleons to become one colour?
Numbers 1 and 2 are written on a whiteboard. Every day Louise’s friend Zara changes these numbers to their arithmetic mean \(a_m\) and harmonic mean \(h_m\).
(The arithmetic mean of two numbers \(a\) and \(b\) is \(a_m=\frac{a+b}{2}\), and harmonic mean of two numbers \(a\) and \(b\) is \(h_m = \frac{2}{\tfrac{1}{a} + \tfrac{1}{b}}\) ).
(a) At some point Zara wrote \(\frac{941664}{665857}\) as one of the two numbers (it is not known which). What was the other number written on the whiteboard at that time?
(b) Can \(\frac{35}{24}\) be ever written by Zara on the whiteboard?
Using mathematical induction prove that \[1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}.\]
There are \(n\) lines on a plane, and all the lines intersect at exactly one point. Prove that the lines divide the plane into \(2n\) parts.
There are \(n\) lines on a plane, no two lines are parallel, and no three lines cross at one point. Show that those lines dived the plane into \(\frac{n(n+1)}{2}+1\) regions.
In a sequence 2, 6, 12, 20, 30, ... find the number
(a) in the 6th place
(b) in the 2016th place.
Using mathematical induction prove that \[1 +3 +5 +\dots + (2n-1) = n^2.\]
Circles and lines are drawn on the plane. They divide the plane into non-intersecting regions, see the picture below.
Show that it is possible to colour the regions with two colours in such a way that no two regions sharing some length of border are the same colour.
Numbers \(1,2,\dots,n\) are written on a whiteboard. In one go Louise is allowed to wipe out any two numbers \(a\) and \(b\), and write their sum \(a+b\) instead. Louise enjoys erasing the numbers, and continues the procedure until only one number is left on the whiteboard.
What number is it? What if instead of \(a+b\) she writes \(a+b-1\)?
Prove that
(a) \[1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6} n (n+1)(2n+1)\]
(b) \[1^2 + 3^2 + 5^2 + \dots + (2n-1)^2 = \frac{1}{3} n (2n-1)(2n+1).\]