Can you find \(11\) distinct whole numbers whose last digits are all different from each other?
The Chinese remainder theorem is a fundamental result in number theory that allows one to decompose congruence problems to into simpler ones. The theorem says the following.
Suppose that \(m_1,m_2\) are coprime (i.e: they have no prime factors in common) natural numbers and \(a_1,a_2\) are integers. Then there is a unique integer \(x\) in the range \(0\leq x \leq m_1m_2-1\) such that \[x \equiv a_1 \pmod{m_1} \quad \text{ and } \quad x \equiv a_2 \pmod{m_2},\] where the notation \(x\equiv y \pmod{z}\) means that \(x-y=kz\) for some integer \(k\). Prove the Chinese remainder theorem using the pigeonhole principle.
We have ten positive integers \(x_1,\dots,x_{10}\) such that \(10\leq x_i\leq 99\) for \(1\leq i\leq 10\). Prove that there are two disjoint subsets of \(x_1,\dots,x_{10}\) with equal sums of their elements.
Look back at Problem 3. You will now prove that actually we can do better! Show that given \(5\) distinct whole numbers - not necessarily consecutive - we can also find three of them such that their sum is divisible by \(3\).
Replace the stars with positive whole numbers so that \[\frac{1}{*}+\frac{1}{*}=\frac{7}{48}.\]
A teacher saw the calculation \(3\times 4 = 10\) written on the whiteboard. She was about to erase it, thinking it was wrong, but then wondered whether it might have been written in a different numeral system.
Is it possible that this multiplication is correct in some base? If so, which one?
On a distant planet called Hexaris, there live two alien species: the Blipnors and the Quantoodles.
The chief alien writes on a board: “There are \(100\) aliens on this planet. Of these, \(24\) are Blipnors and \(32\) are Quantoodles.”
At first this seems confusing — the numbers do not seem to add up! Then you remember that the aliens use a different base for their numeral system.
What base are they using?
Take the numbers \(0,1,2,\dots,3^k-1\), where \(k\) is a whole number.
Show that you can pick \(2^k\) of these numbers so that, among the numbers you picked, no number is the average of two other chosen numbers.
What is the smallest number of weights that allows us to weigh any whole number of grams of gold from \(1\) to \(100\) on a two-pan balance? The weights may be placed only on the left pan.
Now the detective has actually \(126\) people involved in the case! But he is in a big rush and needs to solve the case in only \(9\) days. Can you help him?