Forum Replies Created
AdministratorMarch 21, 2021 at 12:22 pm
That’s beautiful, but there is so much case analysis 😀
Just to let people know, each letter corresponds to a different digit.
AdministratorMarch 15, 2021 at 7:17 pm
This looks like a fun puzzle, I will think about it!
If you don’t mind, I can host the game, so that people can play it without downloading the html file? If the code is yours, let me know so that I add “by Vivek” at the bottom:)
AdministratorMarch 13, 2021 at 12:27 pm
Thanks for the fun problem.
We decompose n = p₁^(a₁)p₂^(a₂)…pₙ^(aₙ) and assume that p₁=n/d(n). Then, we get:
p₁^(a₁ – 1)p₂^(a₂)…pₙ^(aₙ) = (a₁ + 1)(a₂ + 1)…(aₙ + 1).
Now, we note that pᵢ^(aᵢ) ≥ (aᵢ+1) with equality only if pᵢ = 2 and aᵢ = 1.
Then, we must have:
- a₁ = 3 ⇒ n = 8; (easy)
- a₁ = 2 ⇒ n = 9, 12, 18; (easy)
- a₁ = 1 ⇒ n = p₁m, p₁ > 2; Then, aᵢ + 1=2 and 2/n. p₂ = 2 and we get n = 8p, p > 2 or n = 12p, p > 3.
I just brushed over the case analysis and I hope I haven’t missed some solutions… Even though I suspect I have 😀
- a₁ = 3 ⇒ n = 8; (easy)
AdministratorApril 22, 2021 at 4:39 pm
Hey toby, I just sent it to you:)
AdministratorApril 16, 2021 at 8:31 am
Hello Hayden. Can you share your work so that I give more tips?
AdministratorApril 16, 2021 at 8:30 am
Hey Etan, I am sending it now:)
AdministratorApril 4, 2021 at 10:51 pm
Hey Paula, I sent you some hints over PM:)
AdministratorMarch 31, 2021 at 4:05 pm
Hey Paula, feel free to share your work either here or with me via PM and I will help.
AdministratorMarch 31, 2021 at 4:03 pm
Hey Pooh, probably you are overthinking it. If you show me your work, I can help:)
AdministratorMarch 24, 2021 at 10:36 pm
Hey, I am out of hints already 😀 If you would like me to send you the answer via PM, let me know.
AdministratorMarch 22, 2021 at 4:08 pm
That’s right. One of the paintings on the right wall is upside down, that’s a clue:)
AdministratorMarch 20, 2021 at 2:57 pm
I think your mistake is that you have counted the 5 cent coin as a 50 cent coin in the newspaper chapter:)
AdministratorMarch 14, 2021 at 7:22 pm
It is indeed a bit confusing, but I believe all moustaches are black. There is one guy without black on his face and I think it is him:)
AdministratorMarch 13, 2021 at 3:38 pm
Oops, thanks, fixed:)
AdministratorMarch 13, 2021 at 3:37 pm
You are right… My idea for the modification was to pick the sequence x(i), such that x(i+1) ≤ x(i)+1/f(x(i)) and f(x(i+1)) ≥ 2f(x(i)), since that’s all we need for the computation. I see now that:
- The sequence x(i+1)=x(i)+1/f(x(i)) satisfies this condition.
- We are actually FORCED to pick x(i+1)=x(i)+1/f(x(i)), since in the worst case we can get f(x(i)+1/f(x(i)) ≥ 2f(x(i)) and f(y) < 2f(x(i)) for y < x(i)+1/f(x(i).
So, your solution is actually the only possible modification of my solution that works for discontinuous functions 😀
AdministratorMarch 13, 2021 at 9:22 am
You already told me over PM that you have figured things out, but I will leave this here in case someone ever reads it:)
If you pick f(x) = x³ and g(x) = 1/x, then we have:
lim (f(x+g(x)) – f(x))/g(x) – f'(x) = lim x(x³+3x+3/x+1/x³-x³) – 3x² = 3
The example is not perfect, but it illustrates why you can’t replace (f(x+g(x)) – f(x))/g(x) with f'(x) even when g(x) ⇾ 0.
AdministratorMarch 13, 2021 at 9:11 am
That’s right, great solution! A bit of analysis… The key difference between the two solutions is that for my sequence I pick the x‘s for which the value f(x) gets doubled every time. These x‘s are smaller than yours; you have chosen them such that x(i+1) = x(i) + 1/f(x(i)) (which seems a bit more natural). However, in both cases the sequences x(i) converge, which brings a contradiction since f(x) can’t have a vertical asymptote.
It is nice to have both approaches here, thank you for sharing! 🙂
AdministratorMarch 10, 2021 at 1:01 pm
The issue is that the function even doesn’t need to be continuous, which is weaker than differentiability…
AdministratorMarch 10, 2021 at 12:58 pm
I believe the function even doesn’t need to be continuous; only non-decreasing and positive is enough.
Regarding the second point, I agree that 1/f(x) approaches 0 as x grows, but the issue is that f'(x) = lim (f(x+h)-f(x))/h with limit over h for fixed x. In our case we are taking the limit over x as it grows to inifinity, so we can’t quite say that the expression approaches f'(x). In order to see exactly what is happening, we can use the Mean Value Theorem and for every x replace (f(x+1/f(x))-f(x))/(1/f(x)) with f'(x+g(x)), where g(x) < 1/f(x). Then, we will get f'(x+g(x))/f(x)-f(x), but we can’t dismiss g(x), even though it goes to 0.
I can elaborate more on this and probably come up with a counter-example, as it is quite a tricky point. Let me know and I will be happy to do it:)