The Puzzler – Week 3

/ The Weekly Puzzler / By

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.

This week’s shoutout for the Puzzler Quik goes to Gautham from Illinois. The shoutout for the Puzzler Think goes to Joseph from Illinois.

Puzzler Quik

There are three bags, each containing two marbles. Bag A contains two white marbles, Bag B contains two black marbles, and Bag C contains one white marble and one black marble. You pick a random bag and take out one marble, which is white. Let the probability that the remaining marble from the same bag is also white be a simplified fraction $$m/n$$ Find $$m+n$$

Submit Answer

Puzzler Think

Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain).

The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go “Aye”, the loot is divided as proposed, as a majority is definitely needed in a revolt.

If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain.

What is the maximum number of coins the captain can keep without risking his life?

Submit Answer

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

Kudos to Gautham from Ilinois for solving this week’s Puzzler Quik!!

For the man to be 100% certain that he will have a pair of black socks, we must look at the worst-case scenario. If the man is extremely unlucky, he will pull all 21 blue socks and 17 red socks, with no black socks, pulled. Since there are only black socks left, the next two socks he will pull will be black, giving him the pair that he wants. Since there is no scenario worse than this, this scenario is representative of the maximum number of socks he must pull from the drawer. Therefore, our answer is $$21+17+2=40$$

Still have questions? Email me at aditya@thepuzzlr.com for a more in-depth explanation.

Puzzler Think

There are many ways to do this problem, but the simplest way is the one below:

The shout-out for the Puzzler Think goes to Joseph from IL. His explanation was stunning. Below is an explanation provided by homeschoolmath.net interfused with my thoughts.

We know that both of the numbers CANNOT BE BELOW 16. This is because then that number would have at least one more multiple in the first 31 numbers. Since that is not allowed, we know that the #s must be greater than 15. Our answer will contain numbers that are either a power of a number or prime. This is because, for any other number greater than 15, there will be We also know that one of them has to be a prime. Otherwise, the previous factors will multiply to make that number, contradicting itself again. Let’s start going up the cases. First, we have 16&17. 16 is a very special number because the previous factors cannot make it up. If we have 2 and 8, the 2 is useless because the 8 already covers it. 17 is prime, and thus we have our answer: 16+17=33.

Clearly, since the two numbers are consecutive, one of them is even and the other is odd. Let’s say the two numbers were 6 and 7. This would mean the secret number was not divisible by 5 nor by 6. It would also mean the number could NOT be divisible by 14, 21, 28 or by 12, 18, 24 or 30. Thus the students who said it was divisible by 12, by 14, etc. would also have spoken wrong. The two numbers, therefore, cannot be 5 and 6 as only two students are allowed to be incorrect.

Similarly, the two numbers cannot be, say, 8 and 9, because then also the students who said it was divisible by 16, 24, 18, and 27 would have been in the wrong.

So we can conclude that these two consecutive numbers cannot have multiples that are less than 31. This is because then that number would have at least one more multiple in the first 31 numbers. Since that is not allowed, we know that the #s must be greater than 15. This eliminates a lot of numbers: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, and 15.

Could it be, say 20 and 21? That will not work, because if this secret number IS divisible by 4 and 10, then it is also divisible by 20. Similarly, since we know the number IS divisible by all the whole numbers from 2 through 15, it must also be divisible by 18, 20, 21, 22, 24, 26, 28, and 30.

This leaves the following number pair: 16 & 17.

Or, you can think of it this way. All prime numbers, except 2, are odd. So, one number should be an odd prime and the other should be the highest power of 2 in the range.

Thus, the required two numbers are 24 = 16 and 17, and the two students who spoke wrongly are 15th and 16th. Therefore, our answer is 15+16=31

Also from Illinois, Eddie provided a much more formal explanation for this problem. Here is his solution that I have not edited to maintain the rigor of the original proof.


Solution: First, note that none of the first 15 students could have spoken incorrectly. To see this, let us suppose that the fourteenth student spoke incorrectly; the number is not divisible by 15. Thus, either the number is also not divisible by 14 or 16. However, from the 29th student, we know that 30 divides evenly into the large number (since the student could not have spoken incorrectly due to our assumption that the 14th student spoke incorrectly); this implies that 15 also divides evenly into the large number, thus forming a contradiction. In general, assuming that the nth student, where n is less than or equal to 14, spoke incorrectly results in a contradiction; here, the ( \(2n+1\))th student (reporting that the number is divisible by \(2n+2\)) would have spoken correctly, therefore implying that the nth student spoke correctly as well. Hence, we can eliminate the first through fourteenth positions as possibilities for speaking incorrectly. Let us take this further. From the deduction above, the 4th and 5th students must have spoken correctly. Thus, the large number is a multiple of 5 and 6. This implies that the number must be a multiple of lcm(5,6) = 30 as well. In general, if the mth and nth students spoke correctly, then the student in position \(lcm(m+1,n+1)\) must have spoken correctly as well. Note that any common multiple of \(m+1 \) and \(n+1\) greater than the least is not excluded as a possibility of an incorrect divisor statement. For example, we know that the 1st and 3rd students must have spoken correctly. Thus, the student in position lcm(2,4) = 4 must have spoken correctly (which we know already), but the student in position 16 (a multiple of 4) need not have, since divisibility by 4 does not guarantee divisibility by 16. Eliminating more possibilities of incorrect divisor statement (including \(18 = lcm(2,9)\) and \(21 = lcm(3,7)\)) then gives that the students in positions 15, 16, 18, 22, 24, 26, 28, and 30 (corresponding to the divisibility statements by 16, 17, 19, 23, 25, 27, 29, 31) still may have spoken incorrectly. However, since we are given that the incorrect students are in consecutive positions, the incorrect students must be the 15th and 16th students. Therefore, our answer is 31.

This problem is part of a wide math topic called Number Theory. If you want to improve in Number Theory, I would recommend the Intro to Number Theory course by Art of Problem Solving. It really helped me get my basics down when I was new to competition math.

Have any questions? Want to send me a puzzle to possibly be the next puzzler? Email me at aditya@thepuzzlr.com