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Dec
4
comment Combinations of 5-digit numbers
Not unless you are assuming all the digits are different.
Dec
4
comment Solve $33(1.207)^x = 47(1.547)^x$
The step you did was wrong. From $ab^c=de^c$ you should get $\log a+c\log b=\log d+c\log e$.
Dec
4
comment Colored Blocks Factorial
This is the number of solutions of $x+y+z=19$ with $0\le x\le14$, $0\le y\le10$, $0\le z\le12$. This kind of problem has been answered here many times over. I'll try to find an example.
Dec
4
comment Solve the equation by using logarithms
Don't let them confuse you! Dividing by $8$ is dividing by $8$, no matter what the other side of the equation looks like. What would you do if it were $8x=93/42$?
Dec
4
revised Solve the equation by using logarithms
formatting
Dec
4
comment Solve the equation by using logarithms
Wait a minute --- you've done all the log stuff, now you're down to $8x=A$, and you can't solve that for $x$?
Dec
4
comment Probability of cards without combinatorics
The factor for the possible arrangements is the number of distinct arrangements of the letters KKJJO. I could tell you how to do that, but I can't tell you how to do it without combinatorics.
Dec
4
revised Probability of cards without combinatorics
typos
Dec
4
answered Extensions of homomorphisms
Dec
4
comment Icecream flavors - combinatorics problem
See en.wikipedia.org/wiki/…
Dec
4
comment Subgroups of $G^n$
This might be a difficult problem. The number of subgroups for the simplest nontrivial (abelian) example $G=C_2$ are tabulated at oeis.org/A006116 and begin $2,5,16,67,374,2825,29212,417199\dots$.
Dec
4
comment Largest subset of { 0, 1, 2, …, n } that has no 3+ element arithmetic progressions?
A better OEIS reference is oeis.org/A003002 --- also, see the discussion at E10 in Guy, Unsolved Problems In Number Theory, 3rd ed.
Dec
4
comment Largest subset of { 0, 1, 2, …, n } that has no 3+ element arithmetic progressions?
@Ross, I think that's a little different. That's an infinite sequence with no 3-term AP, constructed greedily. But it may be that for various $n$ you can do better than an initial segment of the greedy sequence.
Dec
4
comment Conditional convergence of Riemann's $\zeta$'s series
Well, $(1-(1/2^{s-1}))\zeta(s)$ has that property, but as $\zeta$ itself diverges at $s=1$, that would seem to settle the matter of convergence for $\sigma\le1$.
Dec
4
answered How to explain the solution
Dec
4
comment How many permutations $\tau$ on 8 elements are there such that $\tau\circ\tau$ is the identity permutation?
If it's necessary to explain a permutation as shuffling cards, I doubt the phrase "disjoint cycle" is going to mean anything to OP.
Dec
4
comment How many permutations $\tau$ on 8 elements are there such that $\tau\circ\tau$ is the identity permutation?
You didn't understand my question? My question is: is this homework?
Dec
4
comment How many permutations $\tau$ on 8 elements are there such that $\tau\circ\tau$ is the identity permutation?
If this is homework, please add the "homework" tag.
Dec
4
comment Conditional convergence of Riemann's $\zeta$'s series
My apologies --- things are a little more complicated than I thought --- see math.harvard.edu/archive/213b_spring_05/dirichlet_series.pdf
Dec
4
comment Conditional convergence of Riemann's $\zeta$'s series
"They converge to the right of a vertical line, and diverge to the left of it." If they converge conditionally, they converge --- but I wrote they diverge, and that means they diverge, and don't converge, period. "uniform convergence" has nothing to do with conditional convergence, it's another topic altogether. I grant you that this doesn't account for $\sigma=1$, but it does account for $\sigma\lt1$.