• Our Services
    • Math Olympiad Courses
  • Books
  • Blog
  • Alumni
  • Contact Us
  • Login

Text:

info@42points.com
42 Points42 Points
  • Our Services
    • Math Olympiad Training
      • AMC 8 & MATHCOUNTS
      • Proof-Based Preparation – Part 1
      • Proof-Based Preparation – Part 2
      • AMC 10 & AMC 12
      • Junior Math Olympiad
      • Practice of Problem Solving
      • Senior Math Olympiad
  • Books
  • Blog
  • Alumni
  • Contact
  • Log in

Puerto Rico Team Selection Test, 2023. Day 1

April 30, 2023 Math Competitions, Math Olympiads, Math Olympiads Topics

Problem 1

A palindrome is a positive integer number that remains the same when its digits are reversed.

(a) How many palindromes of $2023$ digits have at least $2022$ equal digits?

(b) How many palindromes of $2023$ digits have at least $2021$ equal digits?

Solution

Let $n_{2021}$, $n_{2022}$, $n_{2023}$ be the number of palindromes that have exactly $2021$, $2022$ and $2023$ equal digits respectfully. Notice that the number of palindromes that have at least $2022$ equal digits is

\[ n_{2022}+n_{2023} \]

while the number of palindromes that have at least $2021$ equal digit is

\[ n_{2021}+n_{2022}+n_{2023} \]

For $n_{2023}$ all digits of the palindromes are the same and we have $9$ possible palindromes, i.e. $n_{2023}=9$.

For $n_{2022}$ the palindromes will have $2022$ digits $x$ and one digit $y$, where $x \neq y$. Since $2023$ is odd, then the middle digit is $y$, and we have $9 \cdot 9$ possible palindromes, i.e. $n_{2023}=81$.

For $n_{2022}$ the palindromes will only have $2021$ digits $x$, one digit $y$ and one digit $z$, where $x \neq y$ and $x \neq z$. Since $2023$ is odd, then the middle digit cannot be $y$, because then the digit $z$ would not have a symmetric digit $z$. Similarly, the middle digit cannot be $z$. This implies that the digits $y$ and $z$ are on the symmetric positions and $y=z$. From here we have $1011 \cdot 9 \cdot 9$ possible palindromes, i.e. $n_{2023}=81891$.

Therefore, the answer for the part (a) is

\[ n_{2022}+n_{2023} = 81 + 9 = \boxed{ 90 } \]

and the answer for the part (b) is

\[ n_{2021}+n_{2022}+n_{2023} = 81891 + 81 + 9 = \boxed{ 81981} \]



Problem 2

Consider the semicircle with center $M$ and diameter $AB$. Let $P$ be a point on the semicircle different from $A$ and $B$, and let $Q$ be the midpoint of the arc $AP$. The line through $M$ and parallel to $QP$ intersects the line $PB$ are the point $S$. Prove that the triangle $PMS$ is isosceles.

Solution
Let the radius of the semicircle be $R$ and $\angle MPB = 2a$.

Since $MP=MB=R$, then the triangle $MBP$ is isosceles and $\angle MBP = 2a$ and $\angle PMB = 180-4a$.

Since $Q$ is the midpoint of the arc $AP$, then $AQ=PQ$. Since $MA=MQ=MP=R$, then the triangles $AMQ$ and $PMQ$ are congruent by $Side-Side-Side$. From here the angle $\angle AMQ$ and $\angle PMQ$. are congruent. Since the $\angle AMB = 180^{\circ}$ we have $\angle AMQ = \angle PMQ = 2a$.

From the triangle $PQM$ we have $\angle MPQ = 90^{\circ} – a$. Since $QP \parallel MS$, then $\angle PMS = \angle MPQ = 90^{\circ} – a$.

Now from the triangle $PMS$ we have $\angle MSP = 90^{\circ} – a$ and the triangle $PMS$ is isosceles, as desired.



Problem 3

Let $A=\{-1,0,1,2\}$. Given that

\[ \sum_{i=1}^{2023} { a_i } = 19 \]

\[ \sum_{i=1}^{2023} { a_i^2 } = 99 \]

where $a_i \in A$, for $i=1,2,…,2023$. Find the maximum and minimum values of

\[ M = \sum_{i=1}^{2023} { a_i^3 } \]

Solution
Let $a$, $b$, $c$ and $d$ be the number of terms $a_i$ that are equal to $-1$, $0$, $1$ and $2$, respectfully. Notice that $a+b+c+d=2023$. From the equality

\[ \sum_{i=1}^{2023} { a_i } = 19 \]

we have

\[ -a+c+2d = 19 \]

which implies that

\[ -a+c=19-2d \]

From here the expression for $M$ becomes

\[ M = \sum_{i=1}^{2023} { a_i^3 } = -a + c + 8d = 19+6d \]

Let us show that the minimal value of $M$ is $19$. Indeed

\[ M = 19+6d \geq 19+6(0) = 19 \]

and the value $19$ is reached for $a=40$, $b=1924$, $c=59$, $d=0$.

Let us show that the maximal value of $M$ is $133$. Indeed, for $d \geq 20$ we have

\[ a+c = 99 – 4d \leq 99 – 4(20) = 19 \]

while at the same time

\[ a+c \geq c-a = 2d-19 \geq 2(20) – 19 = 21 \]

which is impossible. For $d \leq 19$ we have

\[ M = 19+6d \leq 19+6(19) = 133 \]

and the value $133$ is reached for $a=21$, $b=1981$, $c=2$, $d=19$.



Problem 4

Find all the positive integers $n$ that have at least $4$ distinct divisors and $n$ is equal to the sum of the squares of its $4$ smallest positive divisors.

Solution

Answer: $n = \boxed{130}$.

Let

\[ n=a^2+b^2+c^2+d^2 \]
where $a$, $b$, $c$, $d$ are the four smallest divisors of $n$. Start by noticing that $a=1$.

If $n$ is odd, the numbers $b,c,d$ are also odd, and then $a^2+b^2+c^2+d^2$ is even, as we obtained a contradiction.

If $n$ is even, then $b=2$. If $4|n$, then $c=4$ or $d=4$. Since a square of a number is always congruent to $0$ or $1$ mod $4$, then

\[ 0 \equiv n = a^2+b^2+c^2+d^2 = 1 + 4 + 16 + x^2 \equiv 1 + x^2 \pmod{4} \]

which is impossible.

From here $n$ is of the form $4k+2$. This implies that $c$ is an odd prime, and $d=2c$. Therefore, the equation becomes

\[ n=5+5c^2 \]

Notice that the left-hand side of this equation is divisible by $c$. Therefore, the right-hand side should also be divisible by $c$, and therefore $5$ is divisible by $c$. This implies that $c=5$ and $n=130$.


Go to day 2



Share
4

About 42 Points

42 Points is an Online Math Training Program and Tutoring Service. Learn more about our services at https://42points.com/

You also might be interested in

Mathematical Induction: Algebra

Jun 15, 2021

Mathematical Induction is used to prove that the statement of[...]

Vieta’s Formulas

Aug 16, 2021

Let $x_1$ and $x_2$ be the solutions of the quadratic[...]

OMPR 3d Round 2022, High School

Apr 4, 2022

Problem 1 How many sets of integers greater than $1$,[...]

Join our newsletter

Post Archives

Post Categories

Most Liked Posts

  • Solutions to the Polish Mathematical Olympiad, 2021 By 42 Points on December 21, 2021 9
  • Monovariant By 42 Points on June 12, 2021 7
  • Puerto Rico Team Selection Test, 2021. Day 2 By 42 Points on September 14, 2021 7

Tag Cloud

Math Competitions Math Olympiads Math Topics OMPR OMPR 2022 Puerto Rico

Find us on

Ads

Cute Watercolor Bunny Throw Pillow
Adorable Watercolor Bunny Throw Pillow
by ULA Art Studio
Funny Realistic Corn Pattern Socks
Funny Corn Pattern Socks
by ULA Art Studio

42 Points

42 Points is an Online Math Olympiad Program and tutoring service.

Hablamos español, contáctenos para mayor información sobre nuestros cursos y servicios.

 

Contact Information

  • 42 Points
  • info@42points.com
  • 42points.com

Quick Links

  • Math Olympiad Courses
  • AP Calculus
  • Online Math Tutoring
  • Books
  • Blog
  • Alumni
  • Help Center

Fresh from 42Pedia blog

  • Puerto Rico Team Selection Test, 2023. Day 2
  • Puerto Rico Team Selection Test, 2023. Day 1
  • Team Selection Test for Centro and Ibero 2022

WE ACCEPT

PayPal Acceptance Mark

© 2025 — 42 Points.

  • Online Math Training & Tutoring Services
  • Disclaimer
  • Contact
  • Buy AMC 10 Preparation Book
Prev Next