MathDB
Problems
Contests
National and Regional Contests
Serbia Contests
Serbia Team Selection Test
1999 Yugoslav Team Selection Test
1999 Yugoslav Team Selection Test
Part of
Serbia Team Selection Test
Subcontests
(3)
Problem 4
1
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numbers not the sum of 2+ terms in an arithmetic sequence
For a natural number
d
d
d
,
M
d
M_d
M
d
denotes the set of natural numbers which are not representable as the sum of at least two consecutive terms of an arithmetic progression with the common difference d whose terms are integers. Prove that each
c
∈
M
3
c\in M_3
c
∈
M
3
can be written in the form
c
=
a
b
c=ab
c
=
ab
, where
a
∈
M
1
a\in M_1
a
∈
M
1
and
b
∈
M
2
∖
{
2
}
b\in M_2\setminus\{2\}
b
∈
M
2
∖
{
2
}
.
Problem 3
1
Hide problems
permutation of sequence with conditions, existence
Consider the set
A
n
=
{
x
1
,
x
2
,
…
,
x
n
,
y
1
,
y
2
,
…
,
y
n
}
A_n=\{x_1,x_2,\ldots,x_n,y_1,y_2,\ldots,y_n\}
A
n
=
{
x
1
,
x
2
,
…
,
x
n
,
y
1
,
y
2
,
…
,
y
n
}
of
2
n
2n
2
n
variables. How many permutations of set
A
n
A_n
A
n
are there for which it is possible to assign real values from the interval
(
0
,
1
)
(0,1)
(
0
,
1
)
to the
2
n
2n
2
n
variables so that: (i)
x
i
+
y
i
=
1
x_i+y_i=1
x
i
+
y
i
=
1
for each
i
i
i
; (ii)
x
1
<
x
2
<
…
<
x
n
x_1<x_2<\ldots<x_n
x
1
<
x
2
<
…
<
x
n
; (iii) the
2
n
2n
2
n
terms of the permutation form a strictly increasing sequence?
Problem 1
1
Hide problems
polynomial of 2n-th degree
For a natural number
n
n
n
, let
P
(
x
)
P(x)
P
(
x
)
be the polynomial of
2
n
2n
2
n
−th degree such that:
P
(
0
)
=
1
P(0) = 1
P
(
0
)
=
1
and
P
(
k
)
=
2
k
−
1
P(k) = 2^{k-1}
P
(
k
)
=
2
k
−
1
for
k
=
1
,
2
,
.
.
.
,
2
n
k = 1, 2, . . . , 2n
k
=
1
,
2
,
...
,
2
n
. Prove that
2
P
(
2
n
+
1
)
−
P
(
2
n
+
2
)
=
1
2P(2n + 1) - P(2n + 2) = 1
2
P
(
2
n
+
1
)
−
P
(
2
n
+
2
)
=
1
. P.S. I tried to prove it by firstly expressing this polynomial using Lagrange interpolation but get bored of computations - it seems like it can be done this way, but I'd like to see more 'clever' solution. :)