Leaderboard

Al, Bert, and Charlie competed in a track and field event in which points were awarded for 1st, 2nd, and 3rd, place only. At the end of the day, Al had accumulated 22 points, while Bert and Charlie each garnered only 9 points. No other competitor earned points. Bert was 1st in the shot put. Who finished 2nd in the javelin throw? This is a Gold star puzzle.

EDIT: For clarity, I use "number" to reference an individual digit. I use "sequence" to reference the string of numbers. EDIT2: Added spoiler tag.

Perhaps your question may hinge on the source of variability. Pistol has 5-7. In any skirmish, is the choice between 5, 6, 7 made by the program randomly? Or does it depend upon distance to target? Or visibility ( like heavy, light or no fog)? And, as Pickett points out, damage radius is a key parameter of the Dupuy lethality index.

Another factor to possibly consider...

Depends...

Fuzzy thoughts:

You are shown a pile of dimes all of which have one of two distinct weights differing by a small amount not detectable by feel. Forty eight dimes are separated from this pile and you are told of these forty eight, light ones are a dime a dozen (literally - i.e. 44 heavy dimes and 4 light dimes). Using a balance scale twice, find seven heavy dimes. EDIT: for clarification

4 light ones/48

Ten years ago I called attention to a number that when divided by a single integer p it left a remainder of p-1. (Help, a remainder is chasing me) Here is a chance to construct a nine-digit number, a permutation of { 1 2 3 4 5 6 7 8 9 } that has no remainders, sort of. The task is to permute { 1 2 3 4 5 6 7 8 9 } to create a number whose first n digits is a multiple of n for any single-digit n. For example, consider 123654987. Its first 2 digits (12) are divisible by 2. It's first 5 digits (12365) are divisible by 5. However this is not a solution, since 1236549 is not a multiple of 7.

Does the Indy trap appear even if it doesn't save anyone from actual death? Can they act on themselves? Do the goodie save and the Indy trap appear identically in the night post?

Signups: 1. Gavinksong2. Flamebirde3. Nana774. plasmid5. phil18826. bonanova 7. aura8.9. 10.11.12 I've just moved, to Ohio, from NY and drowning in the details of it. Reluctant to pledging my attention here. But, I'll give it a shot ... Should be able to check in at least daily.

If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs, P(BB) = (15/21)×(14/20) = 1/2. The next such arrangement, for which there is exactly 50% chance of taking two blue discs at random, is a box containing eighty-five blue discs and thirty-five red discs. By finding the first arrangement to contain over 1012 = 1,000,000,000,000 discs in total, determine the number of blue discs that the box would contain.

Conjoined twins?

Sorry, something got lost in copying and pasting. The site does not like the formatting used in word, so I am attaching a picture of the solution ... apologies in advance.

import java.util.*; public class lychrel { public static long palin(long n) { long m,p=0; while(n>0) { m=n%10; p=(p*10)+m; n=n/10; } return p; } public static void main(String ar[]) { long a,l,i,j=0,t,it=0; Scanner v=new Scanner(System.in); a=v.nextInt(); l=a+palin(a); t=palin(l); for(i=1;i<=10;i++) { it++; if(l==t) break; else { l=l+palin(l); t=palin(l); } if(i==4) j=l; } if(l!=t) { System.out.println(a+" is a Lychrel Number"); System.out.println("5th iteration of number "+a+" is "+j); } else System.out.println(it); } }

Answer:

Around World Series time recently, (that's the American baseball championships for our international Denizens) I asked the question, dressed as a probability puzzle, "Given that you could pick one of the best-of-seven games in the series that you could guaranty to win, which game would that be? I even flavored the question with the opinions of experts that the pivotal games are often taken to be game 3 or game 5, as statistically these games correlated well with the eventual winner. And many responded with one of those choices for their answers. But the correct answer of course is game 7. (Or as some said, the Last game.) So I also like the "one with the car" choice.

I think phaze is right about the most practical answer. If that fails, I tend to agree with bonanova.

Agree. When I hit the send button, I realized my thinking was too simple. But instead of deleting my post (moderator privilege) I left it to take its licks.

I believe you replace the ? with digits.

Also: http://projecteuler.net/problem=55

Code: #include <iostream> #define BUF_SIZE 100 using namespace std; void putInArray(int num, int* buf){ for(int i=0;i<BUF_SIZE;i++){ buf[i] = num % 10; num = num / 10; } } bool isPalindrom(int* buf){ int i=0, j=BUF_SIZE-1; while(buf[j]==0 && j>0){ j--; } while(i<j){ if(buf[i] != buf[j]){ return false; } i++; j--; } return true; } void applyIteration(int* buf){ int i=0, j=BUF_SIZE-1; while(buf[j]==0 && j>0){ j--; } while(i<=j){ int temp = buf[i] + buf[j]; buf[i++] = temp; buf[j--] = temp; } for(int i=0;i<BUF_SIZE-1;i++){ buf[i+1] += buf[i] / 10; buf[i] = buf[i] % 10; } } bool isLychrel(int* buf){ for(int i=1;i<50;i++){ applyIteration(buf); if(isPalindrom(buf)){ return false; } } return true; } int main() { int buf[BUF_SIZE]; for(int num=0;num<10000;num++){ putInArray(num,buf); if(isLychrel(buf)){ cout << num << " is Lychrel " << endl; } } }

Excellent riddle Plasmid. Those types are my favorite. Give yourself a 5 star vote and make it official. BTW . . . what's the "vote" thing for?

Depends on what is meant by "have." [1] If have means "own," then yes. I can give someone something that I do not own. e.g. if I stole it. To give something, one only needs the ability to determine who controls it. If I control it, I can pass its control to someone else. So ... [2] If have means "possess the control of" then no. As stated, the paradox arises from the different antecedents of "with sorrow." Sorrow is the consequence of giving, not a possession before the act. But the language permits that interpretation by its form. Cute.