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 Oct16 comment 100 sequential parking spaces I don't think we can ignore all the cars above k in your third sentence. For example in the case n=100 if the first car to park happens to be car 3 then we have a much higher chance of parking 3 cars in total than 2 (as cars don't have to park in their own space). Mar18 comment What was the first bit of mathematics that made you realize that math is beautiful? (For children's book) @K.Stm. Ah ok! So |(a,b)| = 2 :-) Mar15 comment What was the first bit of mathematics that made you realize that math is beautiful? (For children's book) @K.Stm. The first point is that the size of any open interval is the same as the size of all the real numbers. This can be seen by stretching a function like tan to create a bijection between the two. You can read more here: en.wikipedia.org/wiki/Cardinality_of_the_continuum Sep10 comment Sequence Sum {1/2 + 1/4 + 1/6 +…} to infinite I think you want four eighths (and correspondingly four terms in the last bracket on the first line). Jul18 comment Divisibility of large number The problem here is the "..." bit. Although 3000, 1500 and 750 are even, and so allow you to use the difference of two squares, 375 is odd, and so there is no next step in this sequence. May24 comment What is the maximum number of edges in a planar bipartite graph which have partitions of order 7 and 15? @DanielPietrobon I thought the two halves of a bipartite graph were called partitions? I'm asking about a bipartite graph with 7 vertices in one half and 15 in the other. Oct27 comment Why can't there be a monotone function with domain $\mathbb{R}$ and range $\mathbb{R} \setminus \mathbb{Q}$? Ah yes - I missed that the question was about hitting all irrationals. Very easy to visualise, thanks! :-) Oct26 comment Why can't there be a monotone function with domain $\mathbb{R}$ and range $\mathbb{R} \setminus \mathbb{Q}$? Just because the function is increasing, doesn't mean that it will reach 0, or 1. Also I don't think you can assume there is a real number y with f(y)=1 if you're also assuming f:R->R\Q. I'm sure you can fix both these issues by a linear transformation of your increasing function (e.g. all increasing functions are a linear transform of an increasing function that has both positive and negative values). Aug23 comment Find thickness of a coin @Mechanical snail: The initial point of contact will almost certainly be one of the four 'corners' shown above. Consequently the centre of mass will almost certainly be to the left/right of the point of contact, and will cause the coin to fall on a head/tail/edge (at which point it will be above the point of contact). Oct4 comment Two question on Permutation and Combination @ Debanjan: I think your list is the list of possible teams, not the list of possible matches (M1,W1)v(M2,W2), (M1,W2)v(M2,W1). Oct4 comment What is the minimal coloring of a planar graph with swaps? Are you interested in all planer graphs with vertices of degree 4, or just the square grid type graph? Sep27 comment Cutting sticks puzzle I am also confused by your suggestion that the number of solutions for the hardest combination is increasing. The link that Chandru1 posted contains a table of the number of solutions for the "hardest" combination. Note that for any even n there is a starting position which has only one solution (up to initial stick reordering and reordering of segments within each initial stick). This is the starting position where there are n/2 sticks of length n+1. Sep27 comment Cutting sticks puzzle I like the idea of solving using OEIS :-) You haven't explained what you are counting as a "solution" above. For example when n=3 there are two possible starting positions: (3,3) and (6). Depending on how you are counting solutions there is either 1 solution {1,2,3} (unordered), or maybe solutions should be counted by where each end stick came from - e.g. (3,3) has a solution where you break 2 off the first stick, and another solution where you break 1 off the second stick, etc. Sep22 comment Connected simple cubic graph +1 Very clean :-) The final inequality should be the other way around: $n \leq 4$. Sep16 comment Grid of overlapping squares Sorry - I have just noticed that I've labelled the grid starting from (1,1) instead of (0,0) as in your question. I've left this as it is because it makes the maths slightly simpler.