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Problems
Contests
International Contests
Francophone Mathematical Olympiad
2023 Francophone Mathematical Olympiad
2023 Francophone Mathematical Olympiad
Part of
Francophone Mathematical Olympiad
Subcontests
(4)
4
2
Hide problems
20n+2 divides 2023n+210
Find all integers
n
⩾
0
n \geqslant 0
n
⩾
0
such that
20
n
+
2
20n+2
20
n
+
2
divides
2023
n
+
210
2023n+210
2023
n
+
210
.
a(a+1)(a+2) ...(a+2023) divides b(b+1)...(b+2023)
Do there exist integers
a
a
a
and
b
b
b
such that none of the numbers
a
,
a
+
1
,
…
,
a
+
2023
,
b
,
b
+
1
,
…
,
b
+
2023
a,a+1,\ldots,a+2023,b,b+1,\ldots,b+2023
a
,
a
+
1
,
…
,
a
+
2023
,
b
,
b
+
1
,
…
,
b
+
2023
divides any of the
4047
4047
4047
other numbers, but
a
(
a
+
1
)
(
a
+
2
)
⋯
(
a
+
2023
)
a(a+1)(a+2)\cdots(a+2023)
a
(
a
+
1
)
(
a
+
2
)
⋯
(
a
+
2023
)
divides
b
(
b
+
1
)
⋯
(
b
+
2023
)
b(b+1)\cdots(b+2023)
b
(
b
+
1
)
⋯
(
b
+
2023
)
?
3
2
Hide problems
2 circles tangent to a third one wanted, one more circle given
Let
Γ
\Gamma
Γ
and
Γ
′
\Gamma'
Γ
′
be two circles with centres
O
O
O
and
O
′
O'
O
′
, such that
O
O
O
belongs to
Γ
′
\Gamma'
Γ
′
. Let
M
M
M
be a point on
Γ
′
\Gamma'
Γ
′
, outside of
Γ
\Gamma
Γ
. The tangents to
Γ
\Gamma
Γ
that go through
M
M
M
touch
Γ
\Gamma
Γ
in two points
A
A
A
and
B
B
B
, and cross
Γ
′
\Gamma'
Γ
′
again in two points
C
C
C
and
D
D
D
. Finally, let
E
E
E
be the crossing point of the lines
A
B
AB
A
B
and
C
D
CD
C
D
. Prove that the circumcircles of the triangles
C
E
O
′
CEO'
CE
O
′
and
D
E
O
′
DEO'
D
E
O
′
are tangent to
Γ
′
\Gamma'
Γ
′
.
2 circles tangent to a third one , IMO SL 2020 variant
Let
A
B
C
D
ABCD
A
BC
D
be a convex quadrilateral, with
∡
A
B
C
>
9
0
∘
\measuredangle ABC > 90^\circ
∡
A
BC
>
9
0
∘
,
∡
C
D
A
>
9
0
∘
\measuredangle CDA > 90^\circ
∡
C
D
A
>
9
0
∘
and
∡
D
A
B
=
∡
B
C
D
\measuredangle DAB = \measuredangle BCD
∡
D
A
B
=
∡
BC
D
. Let
E
E
E
,
F
F
F
and
G
G
G
be the reflections of
A
A
A
with respect to the lines
B
C
BC
BC
,
C
D
CD
C
D
and
D
B
DB
D
B
. Finally, let the line
B
D
BD
B
D
meet the line segment
A
E
AE
A
E
at a point
K
K
K
, and the line segment
A
F
AF
A
F
at a point
L
L
L
. Prove that the circumcircles of the triangles
B
E
K
BEK
BE
K
and
D
F
L
DFL
D
F
L
are tangent to each other at
G
G
G
.
2
2
Hide problems
whatever strategy is the list of numbers on the blackboard remains the same
On her blackboard, Alice has written
n
n
n
integers strictly greater than
1
1
1
. Then, she can, as often as she likes, erase two numbers
a
a
a
and
b
b
b
such that
a
≠
b
a \neq b
a
=
b
, and replace them with
q
q
q
and
q
2
q^2
q
2
, where
q
q
q
is the product of the prime factors of
a
b
ab
ab
(each prime factor is counted only once). For instance, if Alice erases the numbers
4
4
4
and
6
6
6
, the prime factors of
a
b
=
2
3
×
3
ab = 2^3 \times 3
ab
=
2
3
×
3
and
2
2
2
and
3
3
3
, and Alice writes
q
=
6
q = 6
q
=
6
and
q
2
=
36
q^2 =36
q
2
=
36
. Prove that, after some time, and whatever Alice's strategy is, the list of numbers written on the blackboard will never change anymore.Note: The order of the numbers of the list is not important.
Scrooge McDuck owns k gold coins
Let
k
k
k
be a positive integer. Scrooge McDuck owns
k
k
k
gold coins. He also owns infinitely many boxes
B
1
,
B
2
,
B
3
,
…
B_1, B_2, B_3, \ldots
B
1
,
B
2
,
B
3
,
…
Initially, bow
B
1
B_1
B
1
contains one coin, and the
k
−
1
k-1
k
−
1
other coins are on McDuck's table, outside of every box. Then, Scrooge McDuck allows himself to do the following kind of operations, as many times as he likes: - if two consecutive boxes
B
i
B_i
B
i
and
B
i
+
1
B_{i+1}
B
i
+
1
both contain a coin, McDuck can remove the coin contained in box
B
i
+
1
B_{i+1}
B
i
+
1
and put it on his table; - if a box
B
i
B_i
B
i
contains a coin, the box
B
i
+
1
B_{i+1}
B
i
+
1
is empty, and McDuck still has at least one coin on his table, he can take such a coin and put it in box
B
i
+
1
B_{i+1}
B
i
+
1
. As a function of
k
k
k
, which are the integers
n
n
n
for which Scrooge McDuck can put a coin in box
B
n
B_n
B
n
?
1
2
Hide problems
4 b_{n+1} <= P(1)^2
Let
P
(
X
)
=
a
n
X
n
+
a
n
−
1
X
n
−
1
+
⋯
+
a
1
X
+
a
0
P(X) = a_n X^n + a_{n-1} X^{n-1} + \cdots + a_1 X + a_0
P
(
X
)
=
a
n
X
n
+
a
n
−
1
X
n
−
1
+
⋯
+
a
1
X
+
a
0
be a polynomial with real coefficients such that
0
⩽
a
i
⩽
a
0
0 \leqslant a_i \leqslant a_0
0
⩽
a
i
⩽
a
0
for
i
=
1
,
2
,
…
,
n
i = 1, 2, \ldots, n
i
=
1
,
2
,
…
,
n
. Prove that, if
P
(
X
)
2
=
b
2
n
X
2
n
+
b
2
n
−
1
X
2
n
−
1
+
⋯
+
b
n
+
1
X
n
+
1
+
⋯
+
b
1
X
+
b
0
P(X)^2 = b_{2n} X^{2n} + b_{2n-1} X^{2n-1} + \cdots + b_{n+1} X^{n+1} + \cdots + b_1 X + b_0
P
(
X
)
2
=
b
2
n
X
2
n
+
b
2
n
−
1
X
2
n
−
1
+
⋯
+
b
n
+
1
X
n
+
1
+
⋯
+
b
1
X
+
b
0
, then
4
b
n
+
1
⩽
P
(
1
)
2
4 b_{n+1} \leqslant P(1)^2
4
b
n
+
1
⩽
P
(
1
)
2
.
u_{2023} wanted, u_{k+2} >= 2 + u_K, u_{\ell+5} <=5 + u_\ell
Let
u
0
,
u
1
,
u
2
,
…
u_0, u_1, u_2, \ldots
u
0
,
u
1
,
u
2
,
…
be integers such that
u
0
=
100
u_0 = 100
u
0
=
100
;
u
k
+
2
⩾
2
+
u
k
u_{k+2} \geqslant 2 + u_k
u
k
+
2
⩾
2
+
u
k
for all
k
⩾
0
k \geqslant 0
k
⩾
0
; and
u
ℓ
+
5
⩽
5
+
u
ℓ
u_{\ell+5} \leqslant 5 + u_\ell
u
ℓ
+
5
⩽
5
+
u
ℓ
for all
ℓ
⩾
0
\ell \geqslant 0
ℓ
⩾
0
. Find all possible values for the integer
u
2023
u_{2023}
u
2023
.