Topics in Category: Solve One Puzzle A Day - scmGalaxy Forum
http://www.scmgalaxy.com
Thu, 27 Jul 2017 08:39:58 +0000Joomla! - Open Source Content Management/media/kunena/images/icons/rss.pngTopics in Category: Solve One Puzzle A Day - scmGalaxy Forumen-gb4 Jan 2011 - by: aswath.b1@gmail.com
http://www.scmgalaxy.com/forum/60-solve-one-puzzle-a-day/486-4-jan-2011.html#504
http://www.scmgalaxy.com/forum/60-solve-one-puzzle-a-day/486-4-jan-2011.html#504
Let 3X be the number of bullets intially they had..
so after division each will have X bullets.
so after shooting 4 bullets .

X-4+X-4+X-4=X;
2X=12;
X=6;
so the initial number of bullets is 18]]>Solve One Puzzle A DayThu, 06 Jan 2011 17:02:59 +00003 Jan 2011 - Apples and Friends - by: aswath.b1@gmail.com
http://www.scmgalaxy.com/forum/60-solve-one-puzzle-a-day/484-3-jan-2011-apples-and-friends.html#503
http://www.scmgalaxy.com/forum/60-solve-one-puzzle-a-day/484-3-jan-2011-apples-and-friends.html#503
Make one of the friend to hold the basket of apples distribute it to your nine friends and still at the end there is one apple in the basket which you last friend is holding.

Cheers!
Somnath.]]>Solve One Puzzle A DayThu, 06 Jan 2011 17:00:41 +00003 Jan 2011 - by: aswath.b1@gmail.com
http://www.scmgalaxy.com/forum/60-solve-one-puzzle-a-day/485-3-jan-2011.html#502
http://www.scmgalaxy.com/forum/60-solve-one-puzzle-a-day/485-3-jan-2011.html#502
The solution is 7 and 5.

7 on tree A, and 5 on tree B :-)]]>Solve One Puzzle A DayThu, 06 Jan 2011 16:53:32 +00003 Jan 2011 - by: rajeshkumar
http://www.scmgalaxy.com/forum/60-solve-one-puzzle-a-day/481-3-jan-2011.html#483
http://www.scmgalaxy.com/forum/60-solve-one-puzzle-a-day/481-3-jan-2011.html#483
The eldest pirate will propose a 97 : 0 : 1 : 0 : 2 split.

Working backwards, splits in terms of younger to older:

2 Pirates: Pirate Two splits the coins 100 : 0 (giving all to the other pirate). Otherwise, and perhaps even then, Pirate One (the youngest) would vote against him and over he goes!

3 Pirates: Pirate Three splits the coins 0 : 1 : 99. Pirate One (the youngest) is going to vote against him no matter what (see above), but this way, Pirate Two will vote for him, to get at least one gold out of it.

4 Pirates: Pirate Four splits the coins 1 : 2 : 0 : 97. This way, Pirate One will vote for him, and so will Pirate Two - they're getting more than they would under 3 pirates.

5 Pirates: Pirate five splits the coins 2 : 0 : 1: 0 : 97. This way, Pirate One will vote for him, and so will Pirate Three - they're both getting better than they would under 4.]]>Solve One Puzzle A DayMon, 03 Jan 2011 13:55:43 +0000December 30 2010 - by: rajeshkumar
http://www.scmgalaxy.com/forum/60-solve-one-puzzle-a-day/472-december-30-2010.html#482
http://www.scmgalaxy.com/forum/60-solve-one-puzzle-a-day/472-december-30-2010.html#482
Preparations: Divide all 9 nickels to 3 sets, each one of them is 3.

Step 1: Take two sets and weigh them and try to found which one is more heavier. if you found one then you the sets which has heavier one.. else the 3 sets contain the heavier nickel.

Ste 2: Once you have the Set which includes heavier nickel. Once again you divide into 3 and each of will contain single unit of nickel. weigh 2 of them if you found than good else 2 nicke will be your counterfeit.]]>Solve One Puzzle A DayMon, 03 Jan 2011 13:40:12 +0000