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10

Taking the tenth roots, you need to compare $3^5, 4^4, 5^3, 6^2$ since if $a > b$, $a^{10} > b^{10}$. These are $243, 256, 125, 36$. Thus the greatest is $4^{40}$


3

Firstly, let us forget about leading zeroes. A simple way of solving such problems is looking at patterns, how many numerals of each kind, and using the formula [Choose numerals for placing] $\times\;$[Permute them] Here there are only two possible patterns, $3-1-1-1 \;of\;a\;kind:\;\left[\binom{10}1\binom93 \right] \times[ \frac{6!}{3!}] =100,800$ $2-2-...


3

Since the answers in the linked question, I still feel are lacking, here I hope to answer in a more colorful way which avoids "division by symmetry" arguments. Begin as you have by breaking it into cases: either a single number appears three times and three other numbers appear once or two numbers appear twice each and two other numbers appear once each. ...


2

The crucial facts for the solution are: You want to compute $27^{1986} \bmod{256}$ $3^{64} \equiv 1 \bmod{256}$ $3\cdot1986 \equiv 6 \bmod 64$ These imply that $27^{1986} = 3^{3\cdot1986} \equiv 3^6 = 729 \equiv 217\bmod{256}$.


2

Take for instance the number $2132$ (base $4$). I can convert it to base $10$ the following way: $2*4^3 + 1*4^2 + 3*4^1 + 2*4^0 = 158$ So that means $2132$ (base $4$) = $158$ (base $10$). Now what if I want to convert the same number, $2132$ (base $4$) to base $6$? Why can't I do the same method? Example: $2*4^3 + 1*4^2 + 3*4^1 + 2*4^0 = ...


1

One can represent finite trees by a Dyck word, that is a word of the context-free language $D^*$ generated by the grammar $S \to (S) + SS + 1$. The following article gives several possible enumerations for $D^*$: Zoltan Kasa, Generating and ranking of Dyck words, Acta Univ. Sapientiae, Informatica, 1, 1 (2009) 109–118 See also this paper: Yu. S. Medvedeva,...


1

Here's an explanation via an application to a conrete example. $$ x = 0.15\ \overbrace{504}\ 540\ 540\ 540\ \ldots \qquad (\text{“540'' repeats.}) $$ Since the repetend has three digits, we multiply by $1000$ by moving the decimal point over three places: $$ \begin{array}{rcc|c|cc|c|c|c|c|c|c|c|c|c} 1000x & = & 1 & 5 & 5 & . & 4 & ...


1

You got the number to decimal, which is good. Now just take it into base $6$. $6^3 = 216$ is greater than $158$ so we need just three digits. $6^2 \times 4$ is $144$, leaving $14$. $6 \times 2$ is $12$, leaving $2$, and we're done: $$2132_4 = 158_{10} = 422_6.$$ Now, if you wanted to work in base $6$, you could, but then you're just converting each ...



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