q. A total of thirteen pennies is put into three piles so
that each pile has a different number of pennies.
Obviously, this can be done in several ways. Here’s the
challenge – what is the fewest possible number of
pennies in the largest pile of any of the possible
combinations?

6 4 3

5 4 4

sorry, Serial is correct

How is Serial correct? Serial's largest pile is 6 and yours (and mine) is 5. The goal is to have fewest number of pennies in the largest pile. 5<6.

each pile has a different number of pennies

Serial (+1)

Got it. Ali logical puzzle sodhnu paryo bro, not trick questions.

Q.
There are 1000 persons in a circle, numbered 1 thru 1000. Going around the circle, every second person is removed from the circle, starting with person number 2, 4, and so on. this is done repeatedly until one person is left. Which number person out of 1000 will that be?

1

1

977

997

How?

Well, i also think its 1 now...First round ma sabai even numbers will get eliminated..so second round ma you start from 3 (alternate) and eliminate all odd numbers and 1 remains till the end...just a guess

Gabbar Singn Bro: 2 round ma game complete hudaina ni ta...
First round: remove all even numbers.
remaining numbers: 1,3,5,7,9,11....999,
secoun round: remove alternate numbers .starting at 1 again.
remaining numbers: 1, 5, 9, 13 , 17....
third round...............

After every round the total number of people is even.
Every round starts from number 1.
From the first round, when the total number of people is even and the starting number is 1, the people on both sides of 1 go out but 1 remain.
Hence, survivor = 1.

Perfect answer by NAS: +2

Here is my version of code:

import java.util.ArrayList;
import java.util.List;


public class jpt1 {
static int toalPeople=1000;
static int result=0;
static int x;
static int count = 1000;
public static void main(String[] args) {
List result= numberLeft(toalPeople);
System.out.println("last remaining person: "+ result);

}

static public List numberLeft(int x){
ArrayList list = new ArrayList();
for (int i = 1; i <= toalPeople; i++) {
list.add(i);

}
System.out.println("list "+list);
while(count>1){
for (int i = 0; i < list.size()-1; i++) {
list.remove(i+1);
}
System.out.println(" Remaining people : "+list.size());
count = list.size();
}
System.out.println();


return list;
}
}
Output:
Remaining people : 500
Remaining people : 250
Remaining people : 125
Remaining people : 63
Remaining people : 32
Remaining people : 16
Remaining people : 8
Remaining people : 4
Remaining people : 2
Remaining people : 1

last remaining person: [1]

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Niral , Obama, Fuckeeto, Gabbar Singh also answered correctly but without solution so (+1) to each.

NAS bro, ur code is awesome as well and complete with user input. Thats the way it should be. I was trying to be fast, still have 2 unnecessary variable to be cleaned up.

Helpjava11, that is wrong, either your programming is wrong (i don't have knowledge about programming) or something else is wrong. If you remove every second person in a circle, it will work until there is even no. like ...1000 to 500 to 250 to 125. If you remove, 2, 4, 6.......... 998, 1000, then no. 1 will be saved. But when you come to round 4th, after you remove 124th, 125th will be saved and no. 1 will be removed. Please write the no. in the circle and do it by yourself, and you will know it :)