All integer numbers between 1 and 121 are written in the cells of a square grid with size 11 by 11. Then the product of the numbers in every row and the product of the numbers in every column are calculated. Is it possible that the set of all 11 column products coincides with the set of all 11 row-products?
SOLUTION
No, it is not possible. There are 13 prime number between 61 and 121. Since there are only 11 rows, two of them, X and Y, appear in the same row. Now that row is divisible by XY, but clearly, no column is divisible by that number.
A challenge is given to 100 people. A hat will be placed on each of their heads, and each hat will have an integer between 1 and 100 written on it (numbers can repeat). Every person will be able to see the hats of the other 99, but not his own. After that, everyone will have to guess what is the number on their hat (without others hearing). If at least one person guesses correctly, they will be awarded 1 million dollars. What strategy should the people come up with in order to optimize their chance of winning?
SOLUTION
Label the people with numbers 1, 2, 3, … , 100. A strategy which ensures 100% success is the following: Person X should sum the numbers on the hats of the other 99 people, then subtract the result from X, and take the residue modulo 100 of the answer (say “100” if the residue is 0). This way if the sum of the numbers on all hats has residue R when divided by 100, then person R will guess correctly the number on his hat.
Pinkbird is trying to get to Redbird across the river. Where should we place the bridge, so that the path between the two birds becomes as short as possible?
Remark: The bridge is exactly as long as the river is wide, and must be placed straight across it. Additionally, it has some positive width.
SOLUTION
Notice that no matter how the bridge gets placed over the river, the shortest path would be to go to its top left corner, then traverse it diagonally, then go from its bottom right corner to Redbird. The second part of the way has fixed length, so we must minimize the first part plus the third part. In order to do that, imagine we place the bridge, so that its top left corner is at the current position of Pinkbird – point A. If the bottom right corner ends up at point C, then we must connect C with the position of Redbird – point B, and wherever the line intersects the bottom shore – point D, that will be the best place for the bottom right corner of the bridge.
Two moms, Sarah and Courtney, are talking to each other.
Sarah: I have two children. What is the probability that both of Sarah’s children are boys?
Courtney: Me too! Do you have any boys? What is the probability that both of Courtney’s children are boys?
Sarah: Yes, I do! What is your younger child? What is the probability that both of Sarah’s children are boys?
Courtney: It is a boy. He is so mischievous! What is the probability that both of Courtney’s children are boys?
Sarah: Is he Sagittarius? Sagittarius boys are known to drive their mothers crazy. I can testify from personal experience. What is the probability that both of Sarah’s children are boys?
Courtney: No, but actually I have the opposite personal experience to yours. What is the probability that both of Courtney’s children are boys?
Sarah: Well, I guess astrology does not always get it right.
Courtney: I assume it does about half of the time.
SOLUTION
The answers are: ~1/4, ~1/4, ~1/3, ~1/2, ~23/47, 1.
Explanation:
Initially, we do not have any information about the children and therefore the chance that both of them boys is 1/2 × 1/2. This applies to the first and the second question.
After Sarah says that she has at least one boy, there are equal possibilities that she has Boy + Boy, Boy + Girl, or Girl + Boy. Therefore, the chance that both children are boys is 1/3.
After Courtney says that her younger child is a boy, the only remaining question is what is the gender of her older child, and therefore the chance is 1/2.
The fifth exchange implies that Sarah has a Sagittarius boy. There are 23 combinations such that both children are boys and at least one of them is Sagittarius. There are 47 combinations such that at least one of the children is a Sagittarius boy. Therefore, the chance that both children are boys is 23/47.
Finally, Courtney says that her younger child, which is a boy, is not Sagittarius, but her personal experience with Sagittarius boys is positive. Therefore, her older child is a Sagittarius boy and the chance is 1.
I give you a pen and paper and ask you to write the numbers from 1 to 100 in succession so that there are no three numbers such that twice the second one is equal to the sum of the first and the third one. The three numbers do not need to be successive in the sequence.
You have 5 minutes, what do you do?
Remark: The sequence 3, 1, 2, 5, 4 works, but the sequence 1, 4, 2, 5, 3 does not because of the numbers 1, 2, and 3.
and keep iterating until you get a sequence with all numbers from 1 to 128. On each step you take the previous sequence, multiply all elements by 2, and then add the same result but with all elements decreased by 1. This will ensure that the first half contains only even numbers and the second half contains only odd numbers. Since the sum of an odd and an even number is not divisible by 2, if some sequence violates the property, then the previous sequence would have violated it as well.
Once you construct a sequence with 128 numbers, simply remove the numbers from 101 to 128 and you are done. To speed up the process, you can reduce the sequence 8, 16, 4, 12, 6, 14, 2, 10, 7, 15, 3, 11, 5, 13, 1, 9 to 8, 4, 12, 6, 2, 10, 7, 3, 11, 5, 13, 1, 9 and then continue the process.
Use exactly two threes (3) and two eights (8) to get the number 24. You can use multiplication (×), division (÷), addition (+), subtraction (-) signs, and brackets. You can not use any advanced arithmetic operations, such as exponential, factorial, etc.
Can you draw uncountable many non-intersecting “8” shapes in the plane (they can be contained in one another)?
SOLUTION
No, you can’t. For each “8” shape you can choose a pair of points with rational coordinates – one in its top loop and one in its bottom loop. Since no two “8” shapes can have the same corresponding pair of rational points, their number should be countable.
A prince decides to get married to the prettiest girl in his kingdom. All 100 available ladies go to the palace and show themselves to the prince one by one. He can either decide to marry the girl in front of him or ask her to leave forever and call the next one in line. Can you find a strategy which will give the prince a chance of 25% to get married to the prettiest girl? Can you find the best strategy?
Remark: Assume that the prince can objectively compare every two girls he has seen.
SOLUTION
A strategy which ensures a chance of 25% is the following: The prince banishes the first 50 girls which enter the palace and then gets married to the first one which is prettier than all of them (if such one arrives). If the prettiest girl in the kingdom is in the second 50, and the second prettiest girl is in the first 50, he will succeed. The chance for this is exactly 25%.
The best strategy is to wait until ~1/e of all girls pass, and then choose the first one which is more beautiful than all of them. This yields a chance of ~37% for succeeding. The proof is coming soon.
Two people play a game of NIM. There are 100 matches on a table, and the players take turns picking 1 to 5 sticks at a time. The person who takes the last stick wins the game. Who has a winning strategy?
SOLUTION
The first person has a winning strategy. First, he takes 4 sticks. Then every time the second player takes X sticks, the first player takes 6 – X sticks.
Two monks are standing on the two sides of a 2-dimensional mountain, at altitude 0. The mountain can have any number of ups and downs, but never drops under altitude 0. Prove that the monks can climb or descend the mountain at the same time on both sides, always staying at the same altitude, until they meet at the same point.
Please note:
This action will also remove this member from your connections and send a report to the site admin.
Please allow a few minutes for this process to complete.
