There is a thread on this forum discussing which problems can be solved with paper and pencil:

viewtopic.php?f=12&t=1982

## Search found 546 matches

- Sat Jul 28, 2018 12:42 pm
- Forum: News, Suggestions, and FAQ
- Topic: Mark problems that can be figured out through math alone
- Replies:
**3** - Views:
**2394**

- Thu Jul 19, 2018 10:53 am
- Forum: News, Suggestions, and FAQ
- Topic: Add problem categories
- Replies:
**1** - Views:
**1927**

### Re: Add problem categories

This has been asked before. Here is a previous answer:

viewtopic.php?f=5&t=3837&p=41793#p41793

viewtopic.php?f=5&t=3837&p=41793#p41793

- Tue Jul 10, 2018 2:49 pm
- Forum: Number
- Topic: an old maths olympiad question
- Replies:
**28** - Views:
**24524**

### Re: an old maths olympiad question

... hence $\sin 4x=\sin 3x$ implies $\sin(\pi-4x)=\sin 3x$ Nicely done. This is the trick that I missed. I thought we would have to go all the way up to $\sin(7x)$, which would be horrible. NOTE: All this in hindsight knowing that n should be 7. To derive it in an Olympiad is close to impossible fo...

- Mon Jul 09, 2018 4:21 pm
- Forum: Number
- Topic: an old maths olympiad question
- Replies:
**28** - Views:
**24524**

### Re: an old maths olympiad question

So people here can't solve an Olympiad question I probably could but I can't be bothered. Spending a couple of minutes here and there during breaks at work is rather different from actually solving it in Olympiad. Olympiad questions are often hard until some specific insight hits you, and it can ta...

- Mon Jul 09, 2018 1:28 pm
- Forum: Number
- Topic: an old maths olympiad question
- Replies:
**28** - Views:
**24524**

### Re: an old maths olympiad question

I can derive a cubic equation that cos(x) must satisfy where x is half the central angle, i.e. x=180/n. It is straightforward to see that: A1A2 = 2sin(x) A1A3 = 2sin(2x) A1A4 = 2sin(3x) Substituting this in the equation, and using the angle sum formulae sin(a+b)=sin(a)cos(b)+sin(b)cos(a) and cos(2a)...

- Mon Jul 09, 2018 10:02 am
- Forum: Number
- Topic: an old maths olympiad question
- Replies:
**28** - Views:
**24524**

### Re: an old maths olympiad question

Far as I know, a "regular" polygon has sides of equal length, call it length S, so: 1/A1A2 = 1/A1A3 + 1/A1A4 is the same as 1/S^2 = 1/S^2 + 1/S^2, the same as 1 = 1 + 1 so No polygon satisfies... I could be missing something but did you say there were 10 questions? A 1 , A 2 , A 3 , A 4 are consecu...

- Sun Jul 08, 2018 7:27 am
- Forum: Number
- Topic: an old maths olympiad question
- Replies:
**28** - Views:
**24524**

### Re: an old maths olympiad question

Whatever the missing question is, its answer probably relies on the fact that that polynomial can be written as (n-2)(n-1)n(n+1)(n+2)(n+3)+3. I have updated ot now. Is it my inexperience that I cannot see how to factor the polynomial? No. I cheated and typed it into Wolfram Alpha. If I were in a co...

- Sat Jul 07, 2018 3:58 pm
- Forum: Number
- Topic: an old maths olympiad question
- Replies:
**28** - Views:
**24524**

- Wed Jul 04, 2018 7:59 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 141
- Replies:
**27** - Views:
**14361**

### Re: Problem 141

I keep getting the same numbers as Erf_Erf, I've tried a few different approaches and keep getting 23261047907 for 10^10. Can I show my numbers to someone to see what I'm missing? That number is too high. The difference to what it should be is a 10-digit square so probably you have one errant entry...

- Thu May 17, 2018 10:14 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 080
- Replies:
**33** - Views:
**12749**

### Re: Problem 80

There is already a thread for problem 080 here.

In it you will find the answer to your question, which is that the first digit in your example is actually 5.

In it you will find the answer to your question, which is that the first digit in your example is actually 5.

- Fri Feb 23, 2018 10:32 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 085
- Replies:
**9** - Views:
**3653**

### Re: Problem 085

Your answer is a near miss - it is a just little over 2000000 - but you can get closer.

Remember that you can also get close to 2000000 by being just under it.

Remember that you can also get close to 2000000 by being just under it.

- Fri Jan 12, 2018 3:51 pm
- Forum: Recreational
- Topic: Winning Lottery with Prime Number!?
- Replies:
**8** - Views:
**12034**

### Re: Winning Lottery with Prime Number!?

There are 15 primes in the range [1,49]. There are 15C7 + 34C1*15C6 = 176605 ways to choose at least 6 of those primes in a drawing of 7 balls. There are 49C7 = 85900584 ways to draw 7 balls. This gives a probability of 176605/85900584 = 0.00205.., which is about 1 in 500. So this is a rare occurren...

- Mon Dec 18, 2017 10:49 am
- Forum: Number
- Topic: Repeating Decimal and Prime Number
- Replies:
**5** - Views:
**9809**

### Re: Repeating Decimal and Prime Number

It's a directly follows fermat's little theorem: 10^(p-1) == 1 (mod p) I feel embarrassed to admit: I could not see / understand why 10^(p-1) == 1 (mod p) is the reason, even after reading about introductory number theory in these day. I encounterd some congruences such as "Wilson's theorem" but ye...

- Sat Oct 28, 2017 11:56 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 223 & 224
- Replies:
**22** - Views:
**11691**

### Re: Problem 223 & 224

I have written 2 pieces of code for problem 223. I get two different answers , for the perimeter limit = 100.000 I get : val1 = 168921 , val2 = 168676 I can't find my code for this problem right now. What you could do is have your computer compare the triangles generated by the two methods until yo...

- Thu Oct 26, 2017 12:57 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 084
- Replies:
**29** - Views:
**12771**

- Wed Oct 11, 2017 9:09 am
- Forum: News, Suggestions, and FAQ
- Topic: Summer 2017 Website Update
- Replies:
**87** - Views:
**33370**

### Re: Summer 2017 Website Update

There is a help page that explains how to format posts, and which includes a section about the code tag:

https://projecteuler.net/about=forum_tags

https://projecteuler.net/about=forum_tags

- Thu Aug 24, 2017 4:16 pm
- Forum: News, Suggestions, and FAQ
- Topic: Summer 2017 Website Update
- Replies:
**87** - Views:
**33370**

### Re: Summer 2017 Website Update

I think the graphs at problems solved statistics are missing. I see that too. More precisely, this page: https://projecteuler.net/problem_analysis is trying to load these missing images: https://projecteuler.net/cache/statistics_problems_graph.png https://projecteuler.net/cache/statistics_problems_...

- Sun Aug 13, 2017 10:44 pm
- Forum: Number Theory
- Topic: what is the minimum number containing all n-digit substrings?
- Replies:
**1** - Views:
**6674**

### Re: what is the minimum number containing all n-digit substrings?

These are called De Bruijn sequences . It is always possible if you start with all $k$-digit decimal strings (there are $10^k$ of them as leading zeros are allowed), to string them together with maximal overlap to create an optimal sequence that is $10^k+k-1$ long. The last $k-1$ digits will be the ...

- Fri Aug 11, 2017 8:27 pm
- Forum: News, Suggestions, and FAQ
- Topic: Summer 2017 Website Update
- Replies:
**87** - Views:
**33370**

### Re: Summer 2017 Website Update

Anyway, the 'Cached at' info and the time stamp overlaps atleast while on phone. Not just on a phone. Here it is in Firefox on a PC: eulerbug.png The "Cache update" text overlaps the time. Something overlaps the time on most of the pages: Statistics, News, Account, Progress, Recent, and Archives pa...

- Wed Aug 02, 2017 2:20 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 317
- Replies:
**38** - Views:
**16516**

### Re: Problem 317

What approximate value of $\pi$ should be used? You need 4 decimals in your solution. - So if you append the next digit of pi, the solution should not change at all. I solved the problem. :) v6ph1, Your suggestion to use 5 decimals in calculations gives an incorrect result . I think you misundersto...