## Forum Replies Created

Viewing 1 - 20 of 174 posts
• ### Puzzle Prime

March 30, 2021 at 11:15 pm

Hi Pooh, did you check the official walkthrough and the hints topic?

• ### Puzzle Prime

March 21, 2021 at 12:22 pm

Gosh Vivek,

That’s beautiful, but there is so much case analysis 😀

Just to let people know, each letter corresponds to a different digit.

• ### Puzzle Prime

March 15, 2021 at 7:17 pm

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:)

• ### Puzzle Prime

March 13, 2021 at 12:27 pm

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)(a+ 1)…(aₙ + 1).

Now, we note that p^(a) ≥ (a+1) with equality only if p = 2 and a = 1.

Then, we must have:

1. a₁ = 3 ⇒ n = 8; (easy)
2. a₁ = 2 ⇒ n = 9, 12, 18; (easy)
3. a₁ = 1 ⇒ n = pm, 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 😀

• ### Puzzle Prime

April 22, 2021 at 4:39 pm

Hey toby, I just sent it to you:)

• ### Puzzle Prime

April 16, 2021 at 8:31 am

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

• ### Puzzle Prime

April 16, 2021 at 8:30 am

Hey Etan, I am sending it now:)

• ### Puzzle Prime

April 4, 2021 at 10:51 pm

Hey Paula, I sent you some hints over PM:)

• ### Puzzle Prime

March 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.

• ### Puzzle Prime

March 31, 2021 at 4:03 pm

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

• ### Puzzle Prime

March 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.

• ### Puzzle Prime

March 22, 2021 at 4:08 pm

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

• ### Puzzle Prime

March 20, 2021 at 2:57 pm

Hello Marina,

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

• ### Puzzle Prime

March 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:)

• ### Puzzle Prime

March 13, 2021 at 3:38 pm

Oops, thanks, fixed:)

• ### Puzzle Prime

March 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:

1. The sequence x(i+1)=x(i)+1/f(x(i)) satisfies this condition.
2. 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 😀

• ### Puzzle Prime

March 13, 2021 at 9:22 am

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/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.

• ### Puzzle Prime

March 13, 2021 at 9:11 am

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! 🙂

• ### Puzzle Prime

March 10, 2021 at 1:01 pm

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