OMPR 3d Round 2022, High School

Problem 1

How many sets of integers greater than $1$, with two or more elements, have the property that the product of their elements is $210$?

Solution

Notice that $210=2\cdot 3\cdot 5\cdot 7$.

There is only one set of $4$ elements: $\{2,3,5,7\}$.

There are $6$ sets of $3$ elements: $\{2\cdot 3,5,7\}$, $\{2\cdot 5,3,7\}$, $\{2\cdot 7,3,5\}$, $\{3\cdot 5,2,7\}$, $\{3\cdot 7,2,5\}$, $\{5\cdot 7,2,3\}$.

There are $7$ sets of $2$ elements: $\{2\cdot 3,5\cdot 7\}$, $\{2\cdot 5,3\cdot 7\}$, $\{2\cdot 7,3\cdot 5\}$, $\{2,3\cdot 5\cdot 7\}$, $\{3,2\cdot 5\cdot 7\}$, $\{5,2\cdot 3\cdot 7\}$, $\{7,2\cdot 3\cdot 5\}$.

Therefore, the number of such sets is $\boxed{{\displaystyle 14}}$

Problem 2

Find the smallest positive integer $x$ such that the sum of its digits is equal to $2022$ and the product of its digits is ${35}^{11}$.

Solution

Notice that ${35}^{11}=5^{11}\cdot 7^{11}$. Therefore, the only possible digits of the number $x$ are $1$, $5$ and $7$. There are exactly $11$ digits $5$, $11$ digits $7$ and the rest $n$ digits are $1$. Then we have

$$5\cdot 11+7\cdot 11+1\cdot n=2022$$

from where $n=1890$. Therefore, using $1890$ digits are $1$, $11$ digits $5$, and $11$ digits $7$ we obtain the smallest number $x$: $\boxed{{\displaystyle 11\dots 1155\dots 5577\dots 77}}$

Problem 3

It is desired to divide the set $\{1,2,\dots ,n\}$ into $12$ subsets, with no elements in common, in such a way that the sum of the elements of each subset is the same. Find the smallest value of $n$ for which this is possible.

Solution

Since the subsets have no elements in common, then there are no two subsets of $1$ element each. Therefore, there can only be at most one subset of one element. Each other subset has at least two elements and, therefore, $n\ge 11\cdot 2+1=23$. For $n=23$, the division of the set $\{1,2,\dots ,23\}$ into $12$ subsets can be done as follows:

$$\{23\},\{1,22\},\{2,21\},\dots ,\{11,12\}$$

Therefore, the smallest value of $n$ is $\boxed{{\displaystyle 23}}$

Problem 4

Find all positive integers $n$, such that the expression

$$\frac{1+11n}{2n–1}$$

is an integer.

Solution

Notice that

$$2(1+11n)-11(2n–1)=13$$

Since $2n–1$ divides $1+11n$, the it should also divide $13$. From here $2n-1$ can only be $1$, $-1$, $13$ and $-13$, which result in the following positive values of $n$: $\boxed{{\displaystyle 1,7}}$

Problem 5

A hexagon that is made up of $6$ congruent right triangles. What is the ratio of the areas of the inner hexagon to the outer hexagon?

Solution

Since the angle of a regular hexagon is ${120}^{\circ }$, then each right triangle has angles ${30}^{\circ }$, ${60}^{\circ }$, ${90}^{\circ }$. If $x$ is the side of the inner hexagon, then $x\sqrt{3}$ is the side of the outer hexagon. The ratio of the areas, therefore, is equal to

$${\left(\frac{x}{x\sqrt{3}}\right)}^2=\boxed{{\displaystyle \frac{1}{3}}}$$

Problem 6

Let $f$ be a function, which satisfies that

$$f(2022+x)=f(2022–x)$$

for every real number $x$. If the equation $f(x)=0$ has exactly four different solutions, determine the value of the sum of those four solutions.

Solution

Let $g(x)=f(2022–x)$. Then

$$g(-x)=g(x)$$

from where $g(x)$ is even and has exactly four different zeros $x_1$, $x_2$, $x_3$, $x_4$. This implies that these zeros are symmetric with respect to the origin and their sum is zero: $x_1+x_2+x_3+x_4=0$. Notice that if $a$ is a zero of $g(x)$, then $2022-a$ is a zero of $f(x)$. Therefore

$$2022-x_1+2022-x_2+2022-x_3+2022-x_4=8088–(x_1+x_2+x_3+x_4)=\boxed{{\displaystyle 8088}}$$

Problem 7

A circle passes through the vertices $C$ and $D$ and is tangent to the side $AB$ of the squares $ABCD$. What is the ratio of the area of the square to the area of the circle?

Solution

Let $O$ be the center of the circle, $E$ be the point of tangency, $F$ be the midpoint of $CD$, the side of the square be $2x$ and the radius of the circle be $R$. Then from the Pythagorean Theorem for the triangle $OCF$:

$$x^2+(2x-R{)}^2=R^2$$

from where $R=\frac{5x}{4}$.

The ratio of the areas, therefore, can be found as

$$\frac{(2x{)}^2}{\pi {\left(\frac{5x}{4}\right)}^2}=\boxed{{\displaystyle \frac{64}{25\pi }}}$$

Problem 8

Let $x_n$ be a sequence of real numbers, such that

$$x_{n+1}=2x_n–x_{n-1}$$

Given that $x_2=3$, $x_5=7$, what is the value of $x_{10}$?

Solution

Let $x_1=a$. Then it can be shown by induction that

$$x_n=3(n-1)–a(n-2)$$

Since $x_5=7$, then $a=\frac{5}{3}$, and $x_{10}=\boxed{{\displaystyle \frac{41}{3}}}$

Problem 9

For positive integers $x$ and $y$, such that

$$\frac{1}{2022}=\frac{1}{x}–\frac{1}{y}$$

what is the smallest possible value for $x+y$?

Solution

Let us get rid of fractions and rewrite the equality as

$$xy=2022(y-x)$$

and factor it as

$$(2022-x)(y+2022)={2022}^2$$

From here $2022-x$ is a factor of ${2022}^2$. The smallest value of $x+y$ is reached for $x=674$, $y=1011$ and $x+y=\boxed{{\displaystyle 1685}}$

Problem 10

Given

$$M=\sum _{n=2}^{2021}\left(\frac{n+1}{(n-1)!+n!+(n+1)!}\right)$$

Find the value of

$$2022!\cdot (M-\frac{1}{2})$$

Solution

Notice that

$$\frac{n+1}{(n-1)!+n!+(n+1)!}=\frac{n+1}{(n-1)!\cdot (1+n+n(n+1))}=$$

$$\frac{n+1}{(n-1)!\cdot (n+1{)}^2}=\frac{1}{(n-1)!\cdot (n+1)}=$$

$$\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=$$

$$\frac{1}{n!}–\frac{1}{(n+1)!}$$

Therefore, the sum of telescopes

$$M=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+\dots +\frac{1}{2021!}-\frac{1}{2022!}=\frac{1}{2}–\frac{1}{2022!}$$

From here

$$2022!\cdot (M-\frac{1}{2})=2022!\cdot \left(\frac{1}{2}–\frac{1}{2022!}–\frac{1}{2}\right)=\boxed{{\displaystyle -1}}$$