MathDB

3

Part of 2015 Saudi Arabia IMO TST

Problems(4)

(a^2 - b^2 + k)/n is an integer.

Source: 2015 Saudi Arabia IMO TST I p3

7/24/2020
Let nn and kk be two positive integers. Prove that if nn is relatively prime with 3030, then there exist two integers aa and bb, each relatively prime with nn, such that a2−b2+kn\frac{a^2 - b^2 + k}{n} is an integer.
Malik Talbi
number theoryInteger
sum a_ia_j(1 - a_ia_j) >= 0 if sum a_1 = sum a_i^2, for a_i>0

Source: 2015 Saudi Arabia IMO TST II p3

7/24/2020
Let a1,a2,...,ana_1, a_2, ...,a_n be positive real numbers such that a1+a2+...+an=a12+a22+...+an2a_1 + a_2 + ... + a_n = a_1^2 + a_2^2 + ... + a_n^2 Prove that ∑1≤i<j≤naiaj(1−aiaj)≥0\sum_{1\le i<j\le n} a_ia_j(1 - a_ia_j) \ge 0 Võ Quốc Bá Cẩn.
inequalitiesalgebran-variable inequality
no of binary sequences S of length 2015

Source: 2015 Saudi Arabia IMO TST III p3

7/24/2020
Find the number of binary sequences SS of length 20152015 such that for any two segments I1,I2I_1, I_2 of SS of the same length, we have • The sum of digits of I1I_1 differs from the sum of digits of I2I_2 by at most 11, • If I1I_1 begins on the left end of S then the sum of digits of I1I_1 is not greater than the sum of digits of I2I_2, • If I2I_2 ends on the right end of S then the sum of digits of I2I_2 is not less than the sum of digits of I1I_1.
Lê Anh Vinh
combinatoricsnumber theorysum of digits
min, max of (x^2+y^2+z^2)(1/x^2+1/y^2+1/z^2) if (x + y + z) (1/x+1/y+1/z)=10

Source: 2015 Saudi Arabia IMO TST IV p3

7/24/2020
Let a,b,ca, b,c be positive real numbers satisfying the condition (x+y+z)(1x+1y+1z)=10(x + y + z) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)= 10 Find the greatest value and the least value of T=(x2+y2+z2)(1x2+1y2+1z2)T = (x^2 + y^2 + z^2) \left(\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}\right) Trần Nam Dũng
inequalitiesalgebraminmax