Search found 53 matches

by dawghaus4
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...
by dawghaus4
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
by dawghaus4
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.
by dawghaus4
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...
by dawghaus4
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
by dawghaus4
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...
by dawghaus4
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...
by dawghaus4
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...
by dawghaus4
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
by dawghaus4
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 ...
by dawghaus4
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...
by dawghaus4
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...
by dawghaus4
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...
by dawghaus4
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...
by dawghaus4
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 $1000000 = \bar{M}$....
by dawghaus4
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...
by dawghaus4
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
by dawghaus4
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)
by dawghaus4
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)
by dawghaus4
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...