Reputation
11,432
Top tag
Next privilege 15,000 Rep.
Protect questions
Badges
2 22 49
Impact
~226k people reached

Apr
20
answered Analog clock with same hands - sometimes one can't tell time
Apr
20
comment Is $2^{\aleph_0} = \aleph_1$?
@MonkeyKing: Indeed, there's a parallel (no pun intended): It was tried to prove it from the other axioms for some time before it was shown to be independent. However I don't think it was ever considered to be self-evident as the fifth postulate was for millennia.
Apr
20
comment How to maximize the minimal amount not payable with the exchange of at most two coins?
@BrianTung: Thank you for your observations. Especially the Fibonacci one is interesting; I now also notice that the smallest non-2-payable amount in those cases is twice the maximal number plus one, which means that the payable amounts are exactly those below the smallest non-2-payable amount (unlike the other solutions which allow payments of some larger values). But of course that also might be a law of small numbers effect.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
(Follow your previous link; I don't know if the "automatically move" would make a new room)
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
Anyway, I got login to chat working, so maybe we should continue there.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
When the hands are on the same position, you can certainly exchange them. Indeed, any solution must be a multiple of $11$ because rotating a valid hands position by $1/11$ of the full circle gives a valid hands position again. $284$ is not a multiple of $11$.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
And why do you think that?
Apr
19
comment Prove $\sum_{i=2}^{n}\frac{1}{(n-1)n}$ = $\frac{(n-1)}{n}$ using induction.
Also, in the definition of $X$ you surely mean $\frac{n-1}{n}$ on the right hand side (and similarly, $\frac{k-1}{k}$ in eq. (1)).
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
For some reason, login to the chat doesn't currently seem to work, despite the test page giving OK for all tests.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
OK, I've found the problem; it's actually quite simple: The equation $t\equiv 144t\ (\mod 1)$ can be rewritten as $143t\equiv 0\ (\mod 1)$, therefore there are $143$ solutions before removing midnight. Therefore you will not be able to tell the time $143-11=132$ times per 12 hours, or $264$ times per day.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
I just notice: If taken literally, at all moments the current time cannot be inferred from the position of the hands, since you cannot tell from the hands whether it is am or pm. ;-)
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
… which is the only one not doubled by the symmetry. Therefore the total number of times (before correction for the case of both hands being at the same position) must be odd; however $144$ is clearly an even number.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
While I don't yet know the right solution, I'm now convinced that $144$ is wrong: Why? Well, it's obvious that (using the time scale of the answer) if $t$ is a solution, then $-t$ is also a solution (it just corresponds to the mirror image of the clock's face). Now there are exactly two times where $t\equiv -t (\mod 1)$, namely $t=0$ (midnight/noon) and $t=\frac12$ (6 o'clock). Now at 6 o'clock, there's no question which time it is, since if the minute hand points down, the hour hand must point in between two numbers, so it can't point at 12. OTOH midnight/noon is a solution, …
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
Yes, it should affect the result. Assuming the $144$ is right (I'm still thinking about this one), it means you have to subtract $11$ instead of $1$ from it, giving $133$ times in $12$ hours, or $266$ times per day.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
There are in total 11 times in 12 hours where both hands are in the same position; in each of those exchanging them trivially gets the same time, but you know the exact time.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
If you mean the moving things which point to the digits on the clock face then hand is the right word.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
With "needles", do you mean the hands?
Apr
19
comment Prove that $\log_a(b)=-\log_b(a)$
Indeed, there are only counterexamples to the claim: There are no real numbers $a$, $b$ such that $\log_ab=-\log_ba$.
Apr
19
comment Prove that $\log_a(b)=-\log_b(a)$
@Arthur: But $1/x\ne -x$
Apr
19
comment Prove that $\log_a(b)=-\log_b(a)$
@corciacandy: No, $b^{-x} = (a^x)^{-x} = a^{-x^2} \ne a$ unless $x=0$ and $a=1$.