4

Part of 1996 All-Russian Olympiad

Problems(2)

two subsets with no fewer than four common elements.

Source: All-Russian Olympiad 1996, Grade 9, First Day, Problem 4

4/18/2013

In the Duma there are 1600 delegates, who have formed 16000 committees of 80 persons each. Prove that one can find two committees having no fewer than four common members.
A. Skopenkov

probabilityexpected valuecombinatorics proposedcombinatoricsProbabilistic Methoddouble countingHi

$a_1 + a_22^k + a_33^k + \dots + a_mm^k = 0$

Source: All-Russian Olympiad 1996, Grade 10, First Day, Problem 4

4/18/2013

Show that if the integers

a_1

;

\dots

a_m

are nonzero and for each

k =0; 1; \dots ;n

(

n < m - 1

a_1 + a_22^k + a_33^k + \dots + a_mm^k = 0

; then the sequence

a_1, \dots, a_m

contains at least

n+1

pairs of consecutive terms having opposite signs.
O. Musin

algebrapolynomialalgebra proposed