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I'm looking at the Cat in the Hat problem. Can anyone explain to me how a simple math logic is used to derive this equation? $\left(\frac{N}{N+1}\right)^{M-1} = \frac{B}{A}.$

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    I will try my best to be "contained" enough the next time .Thanks for the suggestions and editing.2011-09-17

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Beware of sentences like "Using some simple math logics we can ...". First, there's no such thing as "math logics", so that person more likely than not doesn't know what he or she is talking about. Second, what's simple for one person may be not so simple for someone else, so someone telling you that something is simple either hasn't understood this yet, or is trying to impress and/or intimidate you by saying that something is simple when it might not be for you.

In the present case, we have a hierarchy of $M$ levels of cats. With each of the $M-1$ steps down the hierarchy, the cats become more numerous by a factor $N$ and smaller by a factor $N+1$. So since the number of cats on the highest level is $1$ and the height of a cat on the lowest level is also $1$, the number $B$ of cats on the lowest level is $N^{M-1}$ and the height $A$ of the cat on the highest level is $(N+1)^{M-1}$. Taking the ratio of these two yields the equation.

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    Thanks joriki for helping out the 'simple' way :)2011-09-17
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It is simple recursion or induction.

If the number of levels is $M=2$ then the one big cat is of height $H_2=N+1$ and has $N$ cats of height one in its hat.

If the number of levels is $M=3$ then the one big cat is of height $H_3=(N+1)H_2=(N+1)^2$ where and has $N$ cats of height $H_2$ in its hat and so $N^2$ cats of height one.

So in general, with $M$ levels then the one big cat is of height $A=H_M=(N+1)H_{M-1}=(N+1)^{M-1}$, and has $N$ cats of height $H_{M-1}$ in its hat and so $B = N^{M-1}$ cats of height one.

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    Note that my answer is one minute older than this, so the paragraph on claiming that something is simple wasn't specifically referring to this :-)2011-09-16