• Problem 1 (Proposed by Daniel Nieto Pérez )

Let n be a positive integer greater than or equal to 2. Determine the maximum

value that gcd(x, y) + x + y can take, where x, y are integers such that 1 ≤ x ≤ n,
1 ≤ y ≤ n, and x ̸= y.

(gcd(x, y) denotes the greatest common divisor of x and y)

Answer: We will show that the maximum possible value is 2n.

Solution: First, note that if we choose x = n, y = n − 1, then

x + y + gcd(x, y) = n + (n − 1) + gcd(n, n − 1) = n + (n − 1) + 1 = 2n.

Now suppose that 1 ≤ x ≤ n, 1 ≤ y ≤ n, and x ̸= y. Without loss of general-

ity, assume x < y. It follows that

x + y + gcd(x, y) = x + y + gcd(x, y − x) ≤ x + y + (y − x) = 2y ≤ 2n,

where we have used the fact that y − x is a positive integer.

1
• Problem 2 (Proposed by Antonio Laso González )

Antonio writes the numbers 1, 2, ..., 100 on a blackboard. Every minute, he erases
two numbers a and b and writes the sum of the digits of the number a + b. After 99
minutes, only one number remains on the blackboard. What are the possible values
of this final number?

Answer: The possible values for the final number are 1 and 10.

Solution: We will first prove that the only possible values are 1 and 10 through
two claims.

Claim 1 The final number f must satisfy f ≡ 1 (mod 9). Proof: Consider the sum
of the numbers on the blackboard. Initially, this sum is equal to 1 + 2 + · · · + 100 =
100·101
2
= 50 · 101 ≡ 5 · 2 ≡ 10 ≡ 1 (mod 9). Let S(a + b) denote the sum of the
digits of a + b. Since S(a + b) ≡ a + b (mod 9), this sum does not change (mod 9)
at each step. Therefore, the sum of the numbers after 99 minutes, which is f , must
also be congruent to 1 (mod 9).

Claim 2 The final number f must satisfy f ≤ 18. Proof: First, note that the
numbers written on the blackboard are always less than or equal to 100, because if
a ≤ 100 and b ≤ 100, then S(a+b) ≤ S(199) ≤ 19 < 100. Let a, b, c be the numbers
remaining on the blackboard after 97 minutes. Without loss of generality, assume
S(a + b) and c are the numbers remaining on the blackboard after 98 minutes. It
follows that f = S(S(a + b) + c). Since a ≤ 100 and b ≤ 100, we have S(a + b) ≤
S(199) = 19. Thus, S(a + b) + c ≤ 19 + 100 = 119 =⇒ S(f ) ≤ S(99) = 18.

Next, we show that Antonio can choose the numbers such that the final number
is 10. Proof: Set aside the number 45 from the blackboard, which will be used in
the final step. Perform the process arbitrarily with the remaining numbers. Analo-
gously to what was shown earlier, the resulting number (after 98 minutes) will be
either 1 or 10. In either case, we observe that S(1+45) = S(10+45) = 10, obtaining
10 as the final number.

Finally, we show that Antonio can choose the numbers such that the final num-
ber is 1. Proof: Set aside the number 90 from the blackboard, which will be used
in the final step. Use a similar strategy as before to obtain the number 10 with the
other 99 numbers. Finally, the final number will be S(90 + 10) = S(100) = 1.

2
• Problem 3 (Proposed by Javier Andrés Garcı́a Martı́nez y Antonio Laso González )

Gustavo paints the surface

√ of a cube with edge length 1 such that if two points
are at a distance of 2, they are painted in different colors. Determine the mini-
mum number of distinct colors Gustavo could have used.

Answer: The minimum number of colors Gustavo can use is 4.

Solution: First, we choose

√ four corners of the cube such that any two are sep-
arated by a distance of 2. It follows that Gustavo must use a different color for
each corner, i.e., he must use at least 4 colors.

Now we provide a coloring of the √ surface of the cube using four colors such that if
two points are at a distance of 2, they are painted in different colors. We divide
the cube into four regions as shown in the figure, each consisting of a pyramid with
its apex at a corner of the cube and its base formed by the three adjacent corners.
We paint each of these pyramids entirely in one color, with the exception of: -
Points belonging to two regions (those lying on a diagonal of a face), which we paint
arbitrarily in one of the colors of the regions to which they belong; and - Points
belonging to three regions (four of the cube’s corners), which we paint each with
a unique color, ensuring that no two are of the same color and that each corner is
painted with a color matching one of the regions to which it belongs.

Once each point of the cube has been painted, we√verify that the maximum dis-
tance between two points within a single region is 2, which occurs only between
pairs of the four corners of the cube. However, by our coloring, each of these corners
is painted in a different color. Since, in our construction, two points can only share
the same color if they belong to the same region, this coloring satisfies the desired
property.

3
• Problem 4 (Proposed by Javier Badesa Pérez )

Find all integers n for which there exist infinitely many positive integers C, A,
m, a such that Ca − mA = 1 and

(C 2 + 90CA + A2 ) · (m2 + 90ma + a2 ) + n

is a perfect square.

Answer: The only n that satisfies the condition is n = 2024.

Solution: First, note that Ca − mA = 1 =⇒ C 2 a2 + m2 A2 = 1 + 2CAma.

We begin by expanding (C 2 + 90CA + A2 )(m2 + 90ma + a2 ) under the assumption
that C 2 a2 + m2 A2 = 1 + 2CAma:

(C 2 + 90CA + A2 )(m2 + 90ma + a2 ) =

C 2 m2 + A2 a2 + C 2 a2 + A2 m2 + 8100CAma + 90CA(m2 + a2 ) + 90ma(C 2 + A2 ) =

(Cm + Aa)2 + (Ca + Am)2 + 8096CAma + 90(Cm + Aa)(Ca + Am) =
(Cm + Aa + 45(Am + Ca))2 − 2024(Am + Ca)2 + 8096CAma =
(Cm + Aa + 45(Am + Ca))2 − 2024(C 2 a2 + m2 A2 − 2CAma) =
(Cm + Aa + 45(Am + Ca))2 − 2024.

Therefore, if n = 2024, any combination of positive integers C, A, m, a such that

Ca − mA = 1 satisfies the condition, and since there clearly exist infinitely many
such combinations, n = 2024 works.

On the other hand, if n ̸= 2024, then

(Cm + Aa + 45(Am + Ca))2 − 2024 + n = d2 =⇒

(Cm + Aa + 45(Am + Ca))

is bounded, and since C, A, m, a are positive integers, there will only be a finite
number of solutions, as desired.

4
• Problem 5 (Proposed by Antonio Laso González )

Marı́a places three spherical balls b1 , b2 , and b3 inside her spherical bag. The balls
touch the bag at points T1 , T2 , and T3 , respectively. Next, Marı́a places a flat sheet
of paper inside the bag, which rests on top of the three balls at points P1 , P2 , and
P3 , respectively. Prove that the lines T1 P1 , T2 P2 , and T3 P3 pass through a common
point.
Note: The balls b1 , b2 , and b3 are not necessarily tangent to one another.

Solution: We will prove that each line passes through the point on the bag that is
highest with respect to the sheet of paper. Consider the homothety centered at T1
that maps the sphere b1 to the bag, which we denote by Ω. This homothety maps
the sheet of paper to a plane α that is tangent to Ω and parallel to the original
sheet (and located above the sheet). It follows that P1 is mapped to the tangency
point Q of α with Ω. Since the plane α is uniquely determined by the sheet and the
spherical bag Ω (taking into account that it must be above the sheet because the
three balls are below the sheet), it follows that the three lines T1 P1 , T2 P2 , and T3 P3
pass through Q, as desired.

In the following diagram, the argument used in the proof is represented in two dimen-
sions, replacing spheres with circles and planes with lines, and with 2 balls instead
of 3.
Q

P1 P2

T1

T2

Note: It is reasonable to assume that the three balls are mutually tangent to each
other due to the action of gravity. However, this is not relevant to our solution.