MathDB
Problems
Contests
National and Regional Contests
Estonia Contests
Estonia Team Selection Test
2020 Estonia Team Selection Test
2020 Estonia Team Selection Test
Part of
Estonia Team Selection Test
Subcontests
(3)
1
2
Hide problems
(k(n))! lcm (1, 2,..., n)> (n - 1) !
For every positive integer
x
x
x
, let
k
(
x
)
k(x)
k
(
x
)
denote the number of composite numbers that do not exceed
x
x
x
. Find all positive integers
n
n
n
for which
(
k
(
n
)
)
!
(k (n))!
(
k
(
n
))!
lcm
(
1
,
2
,
.
.
.
,
n
)
>
(
n
−
1
)
!
(1, 2,..., n)> (n - 1) !
(
1
,
2
,
...
,
n
)
>
(
n
−
1
)!
.
arithmetic sequence wanted, (i - j) m_k + (j - k) m_i + (k - i) m_j = c ,
Let
a
1
,
a
2
,
.
.
.
a_1, a_2,...
a
1
,
a
2
,
...
a sequence of real numbers. For each positive integer
n
n
n
, we denote
m
n
=
a
1
+
a
2
+
.
.
.
+
a
n
n
m_n =\frac{a_1 + a_2 +... + a_n}{n}
m
n
=
n
a
1
+
a
2
+
...
+
a
n
. It is known that there exists a real number
c
c
c
such that for any different positive integers
i
,
j
,
k
i, j, k
i
,
j
,
k
:
(
i
−
j
)
m
k
+
(
j
−
k
)
m
i
+
(
k
−
i
)
m
j
=
c
(i - j) m_k + (j - k) m_i + (k - i) m_j = c
(
i
−
j
)
m
k
+
(
j
−
k
)
m
i
+
(
k
−
i
)
m
j
=
c
. Prove that the sequence
a
1
,
a
2
,
.
.
a_1, a_2,..
a
1
,
a
2
,
..
is arithmetic
2
4
Show problems
3
3
Hide problems
f(x^3+y^3)=f(x^3)+3x^3f(x)f(y)+3f(x)(f(y))^2+y^6f(y)
Find all functions
f
:
R
→
R
f :R \to R
f
:
R
→
R
such that for all real numbers
x
x
x
and
y
y
y
f
(
x
3
+
y
3
)
=
f
(
x
3
)
+
3
x
3
f
(
x
)
f
(
y
)
+
3
f
(
x
)
(
f
(
y
)
)
2
+
y
6
f
(
y
)
f(x^3+y^3)=f(x^3)+3x^3f(x)f(y)+3f(x)(f(y))^2+y^6f(y)
f
(
x
3
+
y
3
)
=
f
(
x
3
)
+
3
x
3
f
(
x
)
f
(
y
)
+
3
f
(
x
)
(
f
(
y
)
)
2
+
y
6
f
(
y
)
operation, 1 * (1 * (1 * (...* (1 * 1)...))
With expressions containing the symbol
∗
*
∗
, the following transformations can be performed: 1) rewrite the expression in the form
x
∗
(
y
∗
z
)
a
s
(
(
1
∗
x
)
∗
y
)
∗
z
x * (y * z) as ((1 * x) * y) * z
x
∗
(
y
∗
z
)
a
s
((
1
∗
x
)
∗
y
)
∗
z
; 2) rewrite the expression in the form
x
∗
1
x * 1
x
∗
1
as
x
x
x
. Conversions can only be performed with an integer expression, but not with its parts. For example,
(
1
∗
1
)
∗
(
1
∗
1
)
(1 *1) * (1 *1)
(
1
∗
1
)
∗
(
1
∗
1
)
can be rewritten according to the first rule as
(
(
1
∗
(
1
∗
1
)
)
∗
1
)
∗
1
((1 * (1 * 1)) * 1) * 1
((
1
∗
(
1
∗
1
))
∗
1
)
∗
1
(taking
x
=
1
∗
1
x = 1 * 1
x
=
1
∗
1
,
y
=
1
y = 1
y
=
1
and
z
=
1
z = 1
z
=
1
), but not as
1
∗
(
1
∗
1
)
1 * (1 * 1)
1
∗
(
1
∗
1
)
or
(
1
∗
1
)
∗
1
(1* 1) * 1
(
1
∗
1
)
∗
1
(in the last two cases, the second rule would be applied separately to the left or right side
1
∗
1
1 * 1
1
∗
1
). Find all positive integers
n
n
n
for which the expression
1
∗
(
1
∗
(
1
∗
(
.
.
.
∗
(
1
∗
1
)
.
.
.
)
)
⏟
n
u
n
i
t
s
\underbrace{1 * (1 * (1 * (...* (1 * 1)...))}_{n units}
n
u
ni
t
s
1
∗
(
1
∗
(
1
∗
(
...
∗
(
1
∗
1
)
...
))
it is possible to lead to a form in which there is not a single asterisk.Note. The expressions
(
x
∗
y
)
∗
(x * y) *
(
x
∗
y
)
∗
z and
x
∗
(
y
∗
z
)
x * (y * z)
x
∗
(
y
∗
z
)
are considered different, also, in the general case, the expressions
x
∗
y
x * y
x
∗
y
and
y
∗
x
y * x
y
∗
x
are different.
(1 + a^1)(1 + a^2)...(1 + a^{p - 1})\equiv 1 (mod q), if a^p \equiv 1 mod q
The prime numbers
p
p
p
and
q
q
q
and the integer
a
a
a
are chosen such that
p
>
2
p> 2
p
>
2
and
a
≢
1
a \not\equiv 1
a
≡
1
(mod
q
q
q
), but
a
p
≡
1
a^p \equiv 1
a
p
≡
1
(mod
q
q
q
). Prove that
(
1
+
a
1
)
(
1
+
a
2
)
.
.
.
(
1
+
a
p
−
1
)
≡
1
(1 + a^1)(1 + a^2)...(1 + a^{p - 1})\equiv 1
(
1
+
a
1
)
(
1
+
a
2
)
...
(
1
+
a
p
−
1
)
≡
1
(mod
q
q
q
) .