MathDB

1

Part of 1998 Iran MO (3rd Round)

Problems(3)

Every integer occurs in this sequence exactly once

Source: Iran Third Round MO 1998, Exam 1, P1

10/31/2010
Define the sequence (xn)(x_n) by x0=0x_0 = 0 and for all n∈N,n \in \mathbb N, x_n=\begin{cases} x_{n-1} + (3^r - 1)/2,&\mbox{ if } n = 3^{r-1}(3k + 1);\\ x_{n-1} - (3^r + 1)/2, & \mbox{ if } n = 3^{r-1}(3k + 2).\end{cases} where k∈N0,r∈Nk \in \mathbb N_0, r \in \mathbb N. Prove that every integer occurs in this sequence exactly once.
modular arithmeticinductionnumber theory unsolvednumber theory
Find all functions N -> N with three conditions

Source: Iran Third Round MO 1998, Exam 2, P1

7/1/2012
Find all functions f:N→Nf: \mathbb N \to \mathbb N such that for all positive integers m,nm,n,
(i) mf(f(m))=(f(m))2mf(f(m))=\left( f(m) \right)^2, (ii) If gcd⁡(m,n)=d\gcd(m,n)=d, then f(mn)⋅f(d)=d⋅f(m)⋅f(n)f(mn) \cdot f(d)=d \cdot f(m) \cdot f(n), (iii) f(m)=mf(m)=m if and only if m=1m=1.
functionnumber theory proposednumber theory
Table and nuts

Source: Iran Third Round MO 1998, Exam 3, P1

7/1/2012
A one-player game is played on a m×nm \times n table with m×nm \times n nuts. One of the nuts' sides is black, and the other side of them is white. In the beginning of the game, there is one nut in each cell of the table and all nuts have their white side upwards except one cell in one corner of the table which has the black side upwards. In each move, we should remove a nut which has its black side upwards from the table and reverse all nuts in adjacent cells (i.e. the cells which share a common side with the removed nut's cell). Find all pairs (m,n)(m,n) for which we can remove all nuts from the table.
combinatorics proposedcombinatorics