Problems

Age
Difficulty
Found: 640

Prove that for all positive integers \(n\) there exists a partition of the set of positive integers \(k\le2^{n+1}\) into sets \(A\) and \(B\) such that \[\sum_{x\in A}x^i=\sum_{x\in B}x^i\] for all integers \(0\le i\le n\).

Let \(n\) be a positive integer. Show that \(3^{2n+4}-4^n\) is always divisible by \(5\).

Alice and Bob were playing outdoors. A mean lady told them that at least one of them has a muddy face and everyone who has a muddy face must step forward at the same time on the count of three. Then the mean lady will leave them alone.

If a child with clean face steps forward, he is punished. If nobody steps forward, then the mean lady will do the count again. The children are not allowed to signal to each other. How can Alice and Bob avoid punishment?

Alice, Bob and Claire were playing outdoors. A mean lady told them that at least one of them has a muddy face and everyone who has a muddy face must step forward at the same time on the count of three. Then the mean lady will leave them alone.

If a child with clean face steps forward, he is punished. If nobody steps forward, then the mean lady will do the count again. The children are not allowed to signal to each other. How can Alice, Bob and Claire avoid punishment?

Alice, Bob, Claire and Daniel were playing outdoors. A mean lady told them that at least one of them has a muddy face and everyone who has a muddy face must step forward at the same time on the count of three. Then the mean lady will leave them alone.

If a child with clean face steps forward, he is punished. If nobody steps forward, then the mean lady will do the count again. The children are not allowed to signal to each other. How can Alice, Bob, Claire and Daniel avoid punishment?

A group of children were playing outdoors. A mean lady told them that at least one of them has a muddy face and everyone who has a muddy face must step forward at the same time on the count of three. Then the mean lady will leave them alone.

If a child with clean face steps forward, he is punished. If nobody steps forward, then the mean lady will do the count again. The children are not allowed to signal to each other. How can they avoid punishment?

Prove that \(n^{n+1}>(n+1)^n\) for integers \(n\ge3\).

Let \(n\) be a positive integer. Show that \(1+3+3^2+...+3^{n-1}+3^n=\frac{3^{n+1}-1}{2}\).

Show that all integers greater than or equal to \(8\) can be written as a sum of some \(3\)s and \(5\)s. e.g. \(11=3+3+5\). Note that there’s no way to write \(7\) in such a way.

Every year the citizens of the planet “Lotsofteeth" enter a contest to see who has the most teeth.
This year the judge notices:

  1. Nobody has 0 teeth (everyone has at least 1).

  2. There are more people in the contest than the most teeth that any one person has. (For example, if the most teeth anyone has is 27, then there are more than 27 people participating in the contest.)

Must there be two people who have exactly the same number of teeth? Explain why.