MathDB

5

Part of 1994 All-Russian Olympiad

Problems(3)

equality with sums of n variables

Source: All Russian MO 1994 ARO IX P5

7/29/2018
Prove the equality a1a2(a1+a2)+a2a3(a2+a3)+...+ana1(an+a1)=a2a1(a1+a2)+a3a2(a2+a3)+...+a1an(an+a1)\frac{a_1}{a_2(a_1+a_2)}+\frac{a_2}{a_3(a_2+a_3)}+...+\frac{a_n}{a_1(a_n+a_1)}=\frac{a_2}{a_1(a_1+a_2)}+\frac{a_3}{a_2(a_2+a_3)}+...+\frac{a_1}{a_n(a_n+a_1)}
(R. Zhenodarov)
algebraequality
prove [k,m] [m,n] [n,k] >= [k,m,n]^2 for any natural numbers k,m,n

Source: All Russian MO 1994 ARO

7/29/2018
Prove that, for any natural numbers k,m,nk,m,n: [k,m]⋅[m,n]⋅[n,k]≥[k,m,n]2[k,m] \cdot [m,n] \cdot [n,k] \ge [k,m,n]^2
least common multipleInequalitynumber theory
a_{n+1}=a_n+b_n, where b_n is the last digit of a_n & 5 doesn't divide a_n

Source: All Russian MO 1994 ARO

7/29/2018
Let a1a_1 be a natural number not divisible by 55. The sequence a1,a2,a3,...a_1,a_2,a_3, . . . is defined by an+1=an+bna_{n+1} =a_n+b_n, where bnb_n is the last digit of ana_n. Prove that the sequence contains infinitely many powers of two.
(N. Agakhanov)
recursiveSequencepower of 2number theory