MathDB

4

Part of 2001 Estonia National Olympiad

Problems(4)

|x - 1| + |x - 2| +... + |x - 2001| = a exactly one solution

Source: 2001 Estonia National Olympiad Final Round grade 9 p4

3/12/2020
It is known that the equation∣x−1∣+∣x−2∣+...+∣x−2001∣=a |x - 1| + |x - 2| +... + |x - 2001| = a has exactly one solution. Find aa.
algebraequationabsolute value
x^2y^2(x^2+y^2)<=2 if x, y>=0 and x+y= 2

Source: 2001 Estonia National Olympiad Final Round grade 11 p1

3/12/2020
If xx and yy are nonnegative real numbers with x+y=2x+y= 2, show that x2y2(x2+y2)≤2x^2y^2(x^2+y^2)\le 2.
inequalitiesalgebra
no of harmonic triples (a, b, c) is equal to the no of positive divisor of c^2

Source: 2001 Estonia National Olympiad Final Round grade 10 p4

3/12/2020
We call a triple of positive integers (a,b,c)(a, b, c) harmonic if 1a=1b+1c\frac{1}{a}=\frac{1}{b}+\frac{1}{c}. Prove that, for any given positive integer cc, the number of harmonic triples (a,b,c)(a, b, c) is equal to the number of positive divisors of c2c^2.
number theoryharmonicDivisors
for any integer a exists prime p such that 1+a+a^2+...+ a^{p-1} is composite.

Source: 2001 Estonia National Olympiad Final Round grade 12 p4

3/12/2020
Prove that for any integer a>1a > 1 there is a prime pp for which 1+a+a2+...+ap−11+a+a^2+...+ a^{p-1} is composite.
CompositeSum of powersprimenumber theory