Show that the sum of any \(100\) consecutive numbers is a multiple of \(50\) but not a multiple of \(100\).
Alice sums \(n\) consecutive numbers, not necessarily starting from \(1\), where \(n\) is a multiple of four. An example of such a sum is \(5+6+7+8\). Can this sum ever be odd?
Show that the difference between two consecutive square numbers is always odd.
Let \(n\) be a natural number and \(x=2n^2+n\). Prove that the sum of the square of the \(n+1\) consecutive integers starting at \(x\) is the sum of the square of the \(n\) consecutive integers starting at \(x+n+1\).
For example, when \(n=2\), we have \(10^2+11^2+12^2=13^2+14^2\)!
Find the contrapositive of the statement: “If in every school there is a class with at least \(20\) students, then there is a school with at least \(10\) students".
Show that if \(a\) and \(b\) are numbers, then \(a^2-b^2=(a-b)\times (a+b)\).
Show that if \(x,y,z\) are distinct nonzero numbers such that \(x+y+z = 0\), then we have \[\left(\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}\right)\left(\frac{z}{x-y}+\frac{x}{y-z}+\frac{y}{z-x}\right) = 9.\]
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.