Category Archives: Mathematics

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Alice secretly picks two different integers by an unknown process and puts them in two envelopes. Bob chooses one of the two envelopes randomly (with a fair coin toss) and shows you the number in that envelope. Now you must guess whether the number in the other, closed envelope is larger or smaller than the one you have seen.

Is there a strategy which gives you a better than 50% chance of guessing correctly, no matter what procedure Alice used to pick her numbers?

Choose any strictly decreasing function F on the set of all integers which takes values between 0 and 1. Now, if you see the number X in Bob’s envelope, guess with probability F(X) that this number is smaller. If the two numbers in the envelopes are A and B, then your probability of guessing correctly is equal to:

F(A) * 0.5 + (1 – F(B)) * 0.5 = 0.5 + 0.5 * (F(A) – F(B)) > 50%.

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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.

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.

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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.

Start with the following sequences:

1  →  1, 2  →  2, 4, 1, 3  →  4, 8, 2, 6, 3, 7, 1, 5  →  8, 16, 4, 12, 6, 14, 2, 10, 7, 15, 3, 11, 5, 13, 1, 9

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.

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There is a square cake at a birthday party attended by a dozen people. How can the cake be cut into twelve pieces, so that every person gets the same amount of cake, and also the same amount of frosting?

Remark: The decoration of the cake is put aside, nobody eats it.

Divide the boundary of the cake into twelve equal parts, then simply make cuts passing through the separation points and the center. This way all tops and bottoms of the formed pieces will have equal areas, and also all their sides will have equal areas. Since all pieces have the same height, their volumes will be equal as well.

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You have 10 strings of pasta left on your plate. You randomly start tying up their ends, until there are no loose ends anymore. What is the average number of loops which are created?

The expected (average) number of loops at the end of the procedure is equal to the expected number of loops created after the first tying, plus the expected number of loops created after the second tying, etc. After each tying, the number of non-loop strings decreases by 1, and then the probabilities to create a new loop are 1/19, 1/17, 1/15, etc. Therefore, the answer is the sum 1/19 + 1/17 + 1/15 + … + 1/3 + 1/1 ~ 2.1.

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The number 229 has 9 digits, all different. Which digit is missing?

Bonus: Is the number 9991 prime?

Let the missing digit be m. Every number and the sum of its digits give the same remainder when divided by 9. The number 229 = 32 * 644 gives remainder 5 when divided by 9, and therefore 9 divides (0 + 1 + 2 + … + 9) – 5 – m = 40 – m. Thus, the missing digit is 4.

Bonus: 9991 = 10000 – 9 = 1002 – 32 = (100 – 3)(100 + 3) = 97 * 103. Therefore the number 9991 is not prime.

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Borromean rings are rings in the 3-dimensional space, linked in such a way that if you cut any of the three rings, all of them will be unlinked (see the image below). Show that rigid circular Borromean rings cannot exist.

Assume the opposite. Imagine the rings have zero thickness and reposition them in such a way, that two of them, say ring 1 and ring 2, touch each other in two points. These two rings lie either on a sphere or a plane which ring 3 must intersect in four points. However, this is impossible.