Welcome to The Puzzler. Every week, there are 2 new puzzles related to my passion: math, logic, and thinking. The first puzzle will be The Puzzler Quik, meant for those who crave something fun-sized. The second puzzle will be The Puzzler Think, meant for those who love to ponder. The answers will be posted in next week’s column. Don’t forget to submit your answer for a potential shout-out in the next edition of the Puzzler.
Don’t forget to check out our new section: Mind Bogglers. It features counter-intuitive math thoughts and paradoxes.
Unfortunately, despite receiving many submissions for the Puzzler Quik from last week, all of them were incorrect. It was a deceptive puzzler.
The shout out for the Puzzler Think from last week goes to Aakash from San Jose, California.
Puzzler Quik
There are 100 lockers that line the main hallway of Fremd High School. Every night, Mr. Mince, the janitor, closes all the lockers to make sure there is an orderly start to the next day. One day, 100 mischievous students decide that they will play a prank.
The students all meet before school starts and line up. The first student then walks down the hallway and opens every locker. The next student follows by closing every other locker (starting at the second locker). Student 3 then goes to every third locker (starting with the third) and opens it if it’s closed, and closes it if it’s open. Student 4 follows by opening every fourth locker if it’s closed and closing it if it’s open. This goes on and on until Student 100 finally goes to the hundredth locker. When the principal arrives later in the morning, which lockers does she find open?
Puzzler Think
Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:
a. The number should be divisible by 9.
b. If the most right digit is removed, the remaining number should be divisible by 8.
c. If then again the most right digit is removed, the remaining number should be divisible by 7.
d. etc. until the last remaining number of one digit which should be divisible by 1.
Want a hint? Contact me at aditya@thepuzzlr.com
Answers To Last Week’s Puzzlers
Click here to go to last week’s Puzzlers
Puzzler Quik
Last week’s Puzzler Quik was a tough one!
The bag selected is either bag A or bag C because the first marble is white and bag B does not contain a single marble..
If Bag A is selected, the first marble is white and hence the second marble will be white.
If Bag C is selected, the first marble is white and hence the second marble will be black.
Therefore, second marble will be white only if the selected bag is A
Hence, the probability that the remaining marble from the same bag is white
= Probability of selecting Bag A
We know the first marble is white. Bag A has two white marbles whereas bag C has only one white marble. We can choose any one of these 3 white marbles to be the initial marble that we picked. Since there are two white marbles. the probability of selecting Bag A = $$2/3$$
=> Probability that remaining marble from the same bag is white = $$2/3$$
Therefore, our answer is 2+3=5.
We can also solve this problem using casework as there are only 3 cases.
We know that you do not have Bag B (two black marbles) so there are three possibilities
We choose Bag A, the first white marble. The other marble will be white.
We choose Bag A, the second white marble. The other marble will be white.
We choose Bag C, the white marble. The other marble will be black
So 2 out of 3 possibilities are white making our answer 2+3=5.
Interested in learning more about this. There is a more famous version of this problem that is colloquially known as “The Monty Hall Problem”.
Puzzler Think
There are many ways to do this problem, but the simplest way is the one below:
Kudos to Aakash for solving this problem.
I can go on and on explaining this problem and it would still be confusing. That is why I have replaced this week’s explanation with a great video that fully explains the problem. Thanks to Ted-Ed for this video.
Regardless, here is an explanation:
The oldest pirate will propose a 98 : 0 : 1 : 0 : 1 split, in other words the oldest pirate gets 98 coins, the middle pirate gets 1 coin and the youngest gets 1 coin.
Let us name the pirates (from oldest to youngest): Alex, Billy, Colin, Duncan and Eddie.
Working backwards:
2 Pirates: Duncan splits the coins 100 : 0 (giving himself all the gold). His vote (50%) is enough to ensure the deal.
3 Pirates: Colin splits the coins 99 : 0 : 1. Eddie will accept this deal (getting just 1 coin), because he knows that if he rejects the deal there will be only two pirates left, and he gets nothing.
4 Pirates: Billy splits the coins 99 : 0 : 1 : 0. By the same reasoning as before, Duncan will support this deal. Billy would not waste a spare coin on Colin, because Colin knows that if he rejects the proposal, he will pocket 99 coins once Billy is thrown overboard. Billy would also not give a coin to Eddie, because Eddie knows that if he rejects the proposal, he will receive a coin from Colin in the next round anyway.
5 Pirates: Alex splits the coins 98 : 0 : 1 : 0 : 1. By offering a gold coin to Colin (who would otherwise get nothing) he is assured of a deal.
Have any questions? Want to send me a puzzle to possibly be the next puzzler? Email me at aditya@thepuzzlr.com