Found problems: 8
2013 MTRP Senior, 8
Suppose 51 numbers are chosen from 1, 2, 3, ..., 99, 100. Show that there are two such that one divides the other.
2013 MTRP Senior, 6
Let N = {1, 2, . . . , n} be a set of elements called voters. Let C = {S : S $\subseteq$ N} be the power set of N. Members of C are called coalitions. Let f be a function from C to {0, 1}. A coalition S $\subseteq$ N is said to be winning if f(S) = 1; it is said to be losing if f(S) = 0. Such a function is called a voting game if the following conditions hold:
(a) N is a wining coalition.
(b) The empty set $\Phi$ is a losing coalition.
(c) If S is a winning coalition and S $\subseteq$ S' is also winning.
(d) If both S and S' are winning then S $\cap$ S' $\neq$ $\Phi$, i.e S and S' have a
common voter.
Show that the maximum number of winning coalitions of a voting game
is $2^{n-1}$. Also find such a voting game.
2013 MTRP Senior, 4
Let n be an integer such that if d | n then d + 1 | n + 1. Show that n is a prime number.
2013 MTRP Senior, 5
A function f : $R$ $\rightarrow$ $R$ satisfies the property $f(x^2) - f^2(x) \geq 1/4$ for all x.
Verify if the function is one-one.
2013 MTRP Senior, 7
Write 11 numbers on a sheet of paper six zeros and five ones. Perform the following operation 10 times: cross out any two numbers, and if they were equal, write another zero on the board. If they were not equal, write a one. Show that no matter which numbers are chosen at each step, the nal number on the board will be a one.
2013 MTRP Senior, 3
Figure 1 shows a road-map connecting 14 cities. Is there a path passing through each city exactly once?
2013 MTRP Senior, 1
Find how many committees with a chairman can be chosen from a set of n persons. Hence or otherwise prove that
$${n \choose 1} + 2{n \choose 2} + 3{n \choose 3} + ...... + n{n \choose n} = n2^{n-1}$$
2013 MTRP Senior, 2
There are 1000 doors $D_1, D_2, . . . , D_{1000}$ and 1000 persons $P_1, P_2, . . . , P_{1000}$.
Initially all the doors were closed. Person $P_1$ goes and opens all the doors.
Then person $P_2$ closes door $D_2, D_4, . . . , D_{1000}$ and leaves the odd numbered doors open. Next $P_3$ changes the state of every third door, that
is, $D_3, D_6, . . . , D_{999}$ . (For instance, $P_3$ closes the open door $D_3$ and opens
the closed door D6, and so on). Similarly, $P_m$ changes the state of the
the doors $D_m, D_{2m}, D_{3m}, . . . , D_{nm}, . . .$ while leaving the other doors untouched. Finally, $P_{1000}$ opens $D_{1000}$ if it was closed or closes it if it were
open. At the end, how many doors will remain open?