Forum Replies Created

Gosh Vivek,
That’s beautiful, but there is so much case analysis 😀
Just to let people know, each letter corresponds to a different digit.

Hey Vivek,
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:)

Hey Vivek,
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)

Hello Hayden. Can you share your work so that I give more tips?

Hey Paula, feel free to share your work either here or with me via PM and I will help.

Hey Pooh, probably you are overthinking it. If you show me your work, I can help:)

Hey, I am out of hints already 😀 If you would like me to send you the answer via PM, let me know.

That’s right. One of the paintings on the right wall is upside down, that’s a clue:)

Hello Marina,
I think your mistake is that you have counted the 5 cent coin as a 50 cent coin in the newspaper chapter:)

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:)

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 😀

Hey Vivek,
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/x³x³) – 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.

Hey Vivek,
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! 🙂

The issue is that the function even doesn’t need to be continuous, which is weaker than differentiability…

Hey Vivek,
I believe the function even doesn’t need to be continuous; only nondecreasing 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 counterexample, as it is quite a tricky point. Let me know and I will be happy to do it:)