Math Olympiad Problems And Solutions Site

: This is a combination problem, and the number of ways to choose \(5\) people from a group of \(20\) is given by: $ \(inom{20}{5} = rac{20!}{5! imes 15!} = 15504\) $.

: We can write \(1000 = 2^3 imes 5^3\) . The largest integer \(n\) such that \(n!\) divides \(1000\) is \(n = 7\) , since $ \(7! = 2^4 imes 3^2 imes 5 imes 7\) \(, which has more factors of \) 2 \( and \) 5 \( than \) 1000$. Problem 4: Combinatorics A committee of \(5\) people is to be formed from a group of \(10\) men and \(10\) women. How many ways can this be done? math olympiad problems and solutions

: This is a quadratic equation that can be factored as $ \((x+1)^2 = 0\) \(. Therefore, \) x = -1$. Problem 2: Geometry In a triangle \(ABC\) , the lengths of the sides \(AB\) , \(BC\) , and \(CA\) are \(3\) , \(4\) , and \(5\) respectively. Find the area of the triangle. : This is a combination problem, and the

Here are some sample math olympiad problems and solutions: Solve for \(x\) in the equation: $ \(x^2 + 2x + 1 = 0\) $ The largest integer \(n\) such that \(n