Extreme principle/solutions

These are the solutions to the problems on the extreme principle.

1

First, we note that it is impossible for a coin $\omega$ to be tangent to six coins $\omega_1, \omega_2, \hdots, \omega_6$ if each of the $\omega_i$ is bigger than $\omega$ .

Now consider the smallest coin. By the above observation, it is tangent to at most five of the others.

2

Assume that there is such a palindrome. It is required that $n>1=10^0$ . Let $k$ be the maximum positive integer, and $d$ the maximum positive digit, such that $d\cdot10^k\leq n$ . Then, the substring $d\underbrace{00\ldots0}_{k}$ appears in the palindrome. This implies that its reverse, the substring $\underbrace{0\ldots00}_{k}d$ also appears in the palindrome. The substring of $k$ zeros in this, must be entirely part of the decimal representation of one the integers that get appended to form the palindrome. This is because decimal representations of positive integers do no begin with $0$ . That integer must be smaller than $10^{k+1}$ . Therefore it must be of the form $r\cdot10^k$ , with $r\leq d$ , and be followed in the palindrome by a digit $d$ . It cannot be, that $r<d$ , since after $r\underbrace{0\ldots00}_k$ it should be appended $r\underbrace{0\ldots00}_{k-1}1$ . It cannot be that $r=d$ , in other words that the number $d\underbrace{00\ldots0}_k$ is its own reverse inside the palindrome, since before it there should be $\underbrace{99\ldots9}_k$ and after it $d\underbrace{00\ldots0}_{k-1}1$ , which cannot turn into each other when the palindrome is reversed.

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Extreme principle/solutions

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