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7:33 AM
@MichaelHale It's one of the less interesting questions, but I think that one of my favourite WL "killer apps" is the tracking of units like in question 3.
 
 
6 hours later…
1:07 PM
Yeah, that was one of the reasons I wanted to do the test, was use some features that I don't normally use. The Ctrl+= input worked smoothly for all of the unit combinations. But that was my first time using the net access since the 12.1.1 upgrade I guess, so I had to sign in again otherwise it just gave a cryptic no internet access error, despite network test succeeding.
I wish we had more transparency into why like #13 needed so much guidance.
 
 
2 hours later…
2:49 PM
@MichaelHale 12.1.1 was able to solve #13 directly for me, although it did take a few minutes.
 
3:17 PM
Alright. I'll give it an hour. It's been going for 15 min or so I think.
Solve[(((n^(1/c)) n)^(1/b) n)^(1/a) == n^(25/36) && a > 1 && b > 1 &&
  c > 1 && {a, b, c} \[Element] Integers, b]
Oops, forgot constraint on n, I'll restart it.
 
I did:
Solve[Power[n Power[n Power[n, (c)^-1], (b)^-1], (a)^-1] == Power[
n^25, (36)^-1] && a > 1 && b > 1 && c > 1 && n > 1, b,
PositiveIntegers] // AbsoluteTiming
takes about 5 minutes for me
{274.174, {{b ->
ConditionalExpression[3,
n \[Element] Integers && a == 2 && c == 6 && n >= 2]}}}
 
Hmm. Adding Reals at the end of this makes it solve in just a couple of seconds for me.
Solve[(((n^(1/c)) n)^(1/b) n)^(1/a) == n^(25/36) && a > 1 && b > 1 &&
  c > 1 && {a, b, c} \[Element] Integers && n > 1, b, Reals]
But I figured that was redundant given each variable has an inequality associated with it.
 
interesting!
 
And then if I write it with radicals I immediately get a message saying it can't be solved with available methods
Solve[Surd[n Surd[n Surd[n, c], b], a] == Surd[n^25, 36] && a > 1 &&
  b > 1 && c > 1 && {a, b, c} \[Element] Integers && n > 1, b, Reals]
 
I also get ~6s using Reals:
Solve[Power[n Power[n Power[n, (c)^-1], (b)^-1], (a)^-1] == Power[
n^25, (36)^-1] && a > 1 && b > 1 && c > 1 &&
n > 1 && {a, b, c} \[Element] PositiveIntegers, b,
Reals] // AbsoluteTiming
which only makes your point, the system seems to need quite a bit of guidance
I'm going to submit that as an issue
 
3:37 PM
Cool. I was considering the same. I mean, it seems like if it can do it quickly with those little bits of redundant guidance that it should be able to do it quickly as first entered.
#22 had a smooth progression of attempts. First try Sum, then NSum, then just Total@Table
So I at least had a progression of more manual fallbacks to try that is relevant across a variety of problems.
19, 21, and 25 I had no idea on though
 
 
2 hours later…
5:50 PM
Interesting to compare these three
w/y /. Solve[{w/x == 4/3, y/z == 3/2, z/x == 1/6}]
Solve[{w/x == 4/3, y/z == 3/2, z/x == 1/6, w/y == answer}]
Solve[{w/x == 4/3, y/z == 3/2, z/x == 1/6, w/y == answer}, answer]
Last one says no solutions for me. Technically one argument version of Solve isn't documented either. Sent feedback.
 
6:26 PM
would be interesting to see if someone has done a Maple comparison
Also, do I misunderstand how Solve should work, or shouldn't this give me some answer for #19?

Solve[((2^289) + 1)/((2^17) + 1) ==
Sum[2^n, {n, x, 137}], x, NonNegativeIntegers]
 
6:37 PM
I've definitely seen some comparison discussions about Maple with regards to solving integrals. I think Wolfram put in a pretty successful effort to catch up. Haven't seen a Maple run for these problems.
I think you are confusing the sum having 137 terms with the power of the 137th term being 137.
 
7:03 PM
@MichaelHale aha, thank you, reading comprehension problem
 

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