## Search found 53 matches

- Sun Jul 05, 2020 12:08 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 684
- Replies:
**13** - Views:
**3019**

### Re: Problem 684

Could you clarify what you mean about "Instead, s(n) is the inverse digit sum, so digitsum(s(n))=n"? What is an Inverse Digit Sum? Let digitsum(x) be the sum of the digits of the number x. For example, digitsum(19) = 10. An inverse function would reverse that. If the sum of the digits is 10, what i...

- Thu May 21, 2020 1:49 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 077
- Replies:
**8** - Views:
**3655**

### Re: Problem 077

It won't matter which way you want to interpret "sum."

I'm old fashioned, a sum is always two or more numbers, unless otherwise specified.

Tom

I'm old fashioned, a sum is always two or more numbers, unless otherwise specified.

Tom

- Fri Jul 19, 2019 11:58 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 104
- Replies:
**13** - Views:
**5160**

### Re: Problem 104

My program solves the problem in 0.048722 seconds (2.6 GHz Intel core I7). It doesn't use big integers.

- Thu May 10, 2018 5:21 pm
- Forum: Clarifications on Project Euler Problems
- Topic: problem 206
- Replies:
**33** - Views:
**11757**

### Re: Possible error in Euler problem #206

... While i was giving it a go with my calculator and a bit of brainage I found that the number 1 388 659 302 squared gives the number : 1 9 2 8 3 7 4 6 5 5 6 4 7 3 8 2 9 1 0 which to my understanding also respects the form given above ... The square of any number ending in 2 will end in 4. The squ...

- Wed Apr 04, 2018 4:20 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 078
- Replies:
**16** - Views:
**6585**

### Re: Problem 078

The problem doesn't ask for p(n). You need the least n such that p(n) is divisible by 1000000; that is, you need only the last 6 digits of p(n).

Tom

Tom

- Wed Feb 08, 2017 1:02 am
- Forum: Clarifications on Project Euler Problems
- Topic: problem 055
- Replies:
**31** - Views:
**9385**

### Re: problem 055

When a problem "name" a number or concept that I'm not familiar with, I look it up. From Wikipedia: A Lychrel number is a natural number that cannot form a palindrome through the iterative process of repeatedly reversing its digits and adding the resulting numbers. There are times I wish the problem...

- Sun Feb 05, 2017 3:33 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 578
- Replies:
**4** - Views:
**1581**

### Re: Problem 578

The case where k = 0 isn't the assumption that a 0 = 0, but the idea that, if k = 0, then there are no prime products, and we are left with the empty product . See, for example, Wikipedia: This representation is commonly extended to all positive integers, including one, by the convention that the e...

- Sat Feb 04, 2017 9:43 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 578
- Replies:
**4** - Views:
**1581**

### Re: Problem 578

There seems to be an error in the first sentence of the problem. Any positive integer can be written as a product of prime powers: p 1 a 1 × p 2 a 2 × ... × p k a k , where p i are distinct prime integers, a i > 0 and p i < p j if i < j. A positive power of a prime can't equal 1. Tom 1 is the limit...

- Sat Feb 04, 2017 12:31 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 578
- Replies:
**4** - Views:
**1581**

### Problem 578

There seems to be an error in the first sentence of the problem. Any positive integer can be written as a product of prime powers: p 1 a 1 × p 2 a 2 × ... × p k a k , where p i are distinct prime integers, a i > 0 and p i < p j if i < j. A positive power of a prime can't equal 1. Tom

- Tue Jan 24, 2017 12:17 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 090
- Replies:
**46** - Views:
**15141**

### Re: Problem 090

... Am i missing something ? Consider we are able to form all the squares with these two combinations (1) { 0, 1, 2, 6, 7, 9 } { 1, 3, 4, 5, 8, 9 } (2) { 1, 3, 4, 5, 8, 9 } { 0, 1, 2, 6, 7, 9 } the 2nd one is not the duplicate ? I guess earlier post in the forum says its duplicate. If you consider ...

- Mon Jan 09, 2017 3:25 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 129
- Replies:
**13** - Views:
**3978**

### Re: Problem 129

Well, I'm confused because of the request : Find the least value of n for which A(n) first exceeds one-million (10^6) And if we factor R(13) = [53, 79, 265371653], the number 265371653 is > 10^6 . And that's not the answer. But 265371653 is not the least value. The least value is 53. Therefore I mu...

- Sat Jan 07, 2017 1:24 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 103
- Replies:
**38** - Views:
**10797**

### Re: Problem 103

When is the conditional "if p, then q" true? If the hypothesis is false, the conditional is true. (Note: that does not say, when p is false, q is true.) Thus, {1} vacuously satisfies the conditional. Interpret the conditional this way, "Whenever you have two nonempty disjoint subsets, the following...

- Fri Jan 06, 2017 8:04 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 103
- Replies:
**38** - Views:
**10797**

### Re: Problem 103

Hi all, hopefully someone is still checking this thread haha. How can n=1 {1} be a valid optimum special sum set? How can you make two disjoint, non-empty subsets from this? you could have S(B)={1} but then S(C) would have to be empty, unless you can repeat a set number in the subset? Thanks When i...

- Sat Dec 24, 2016 5:02 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 571
- Replies:
**10** - Views:
**2959**

### Re: Problem 571

As far as I know leading zeroes are not allowed on Project Euler since the beginning . In case of doubt, from "The sum of the 10 smallest 10-super-pandigital numbers is 20319792309." you can deduce that the answer should be given in base 10. (just try the computed answer and you will see). So I thi...

- Fri Jul 15, 2016 2:19 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 089
- Replies:
**67** - Views:
**22410**

### Re: Problem 089

You said "I entered 5000 into the roman numeral converter, so it was an invalid number". But, It is written under the converter, in the list of roman numerals. This page written: $5000 = \bar{V}$, $10000 = \bar{X}$, $50000 = \bar{L}$, $100000 = \bar{C}$, $500000 = \bar{D}$, and $1000000 = \bar{M}$....

- Fri Jul 15, 2016 4:55 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 089
- Replies:
**67** - Views:
**22410**

### Re: Problem 089

I think $5000 = \bar{V}$. It's written in this page . So, $4900 = MMMMCM = C \bar{V}$. Is this problem says 4900 can represent only 2 characters? I've never seen a roman numeral character for 5000, not even on the page you linked to. On that page, when I enter 5000 into the pages converter, I'm tol...

- Sat May 28, 2016 2:29 am
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 150
- Replies:
**15** - Views:
**4667**

### Re: Problem 150

I agree with 10 rows, but not 50 and 500 rows.

Tom

Tom

- Wed Feb 17, 2016 9:06 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 127
- Replies:
**36** - Views:
**12245**

### Re: Problem 127

I hope this is fair. Here are my 31 abc-hits with c < 1000. (The sort in the output is on c.)

(snip)

(snip)

- Wed Feb 17, 2016 7:28 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 013
- Replies:
**76** - Views:
**24906**

### Re: Problem 013

I think that if you re-read the forum posts, you will find that it was pointed out that those who thought (snip)

- Fri Feb 12, 2016 2:59 pm
- Forum: Clarifications on Project Euler Problems
- Topic: Problem 015
- Replies:
**63** - Views:
**9443**

### Re: Problem 015

The question shows a 2x2=4 square grid. Am I right to assume that the solution to the 2x2 is based on 3x3=9 sides. If so then can you confirm that the solution to the 20x20 question is based on 21x21 sides. No, that assumption is not needed. There is no need to look at a 21X21 grid to get the numbe...