MathDB

6

Part of 1998 Tournament Of Towns

Problems(4)

TOT 1998 Spring AJ6 10 people, circle table, 100 nuts totally, 10 nuts at end

Source:

5/11/2020
1010 people are sitting at a round table. There are some nuts in front of each of them, 100100 nuts altogether. After a certain signal each person passes some of his nuts to the person sitting to his right . If he has an even number of nuts, he passes half of them; otherwise he passes one nut plus half of the remaining nuts. This procedure is repeated over and over again. Prove that eventually everyone will have exactly 1010 nuts.
(A Shapovalov)
gamegame strategycombinatorics
TOT 1998 Spring AS6 5 cards from a 52-card deck trick

Source:

5/11/2020
(a) Two people perform a card trick. The first performer takes 55 cards from a 5252-card deck (previously shuffled by a member of the audience) , looks at them, and arranges them in a row from left to right: one face down (not necessarily the first one) , the others face up . The second performer guesses correctly the card which is face down. Prove that the performers can agree on a system which always makes this possible. (b) For their second trick, the first performer arranges four cards in a row, face up, the fifth card is kept hidden. Can they still agree on a system which enables the second performer to correctly guess the hidden card?
(G Galperin)
combinatoricsgamegame strategy
TOT 1998 Autumn AJ6 gang of robbers took away a bag of coins

Source:

5/11/2020
A gang of robbers took away a bag of coins from a merchant . Each coin is worth an integer number of pennies. It is known that if any single coin is removed from the bag, then the remaining coins can be divided fairly among the robbers (that is, they all get coins with the same total value in pennies) . Prove that after one coin is removed, the number of the remaining coins is divisible by the number of robbers.
(Folklore, modified by A Shapovalov)
combinatorics
TOT 1998 Autumn AS6 f (x) = f_1 (f_2( ... f_{n-1}(f_n (x) )... ))

Source:

5/11/2020
In a function f(x)=(x2+ax+b)/(x2+cx+d)f (x) = (x^2 + ax + b )/ (x^2 + cx + d) , the quadratics x2+ax+bx^2 + ax + b and x2+cx+dx^2 + cx + d have no common roots. Prove that the next two statements are equivalent: (i) there is a numerical interval without any values of f(x)f(x) , (ii) f(x)f(x) can be represented in the form f(x)=f1(f2(...fn−1(fn(x))...))f (x) = f_1 (f_2( ... f_{n-1} (f_n (x))... )) where each of the functions fjf_j is o f one of the three forms kjx+bj,1/x,x2k_j x + b_j, 1/x, x^2 .
(A Kanel)
quadraticsfunctioncompositionalgebra