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Difference between revisions of "2023 IOQM Problems"

(→‎Solution)
(→‎Problem 9)
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==Problem 9==
 
==Problem 9==
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Find the number of triples <math>(a, b, c)</math> of positive integers such that (a) <math>ab</math> is a prime;
+
Find the number of triples <math>(a, b, c)</math> of positive integers such that
 +
(a) <math>ab</math> is a prime;
  
 
(b) <math>bc</math> is a product of two primes;
 
(b) <math>bc</math> is a product of two primes;
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(d) <math>abc\leq30</math>
 
(d) <math>abc\leq30</math>
 +
==Solution==
 +
ab is a prime means than one of ab is 1.
 +
*Case I: <math>a=1</math> so b is a prime and bc is product of two pimes. Also, <math>bc \le 30</math> and <math>bc\neq {4, 9, 25}</math> and (b,c) are different primes. This gives {<math>b=2,3,5,7</math>} and {<math>c =3,5,7</math>}, on counting the total number of cases comes out to be 14.
 +
*Case II: when b=1 only 3 triplets which will satisfy the above conditions are: (2,1,15), (3,1,10), (5,1,6).
 +
Number of triplets: <math>\boxed{14+3=\textbf{17}}</math>
 +
 
==Problem 10==
 
==Problem 10==
 
The sequence <math>\{a_n\}_{n\geq0}</math> is defined by <math>a_0 = 1</math>, <math>a_1 = -4</math>, and <math>a_{n+2} = -4a_{n+1} - 7a_n, \text{for } n \geq 0.</math> Find the number of positive integer divisors of <math>a_{50}^2 - a_{49}a_{51}</math>.
 
The sequence <math>\{a_n\}_{n\geq0}</math> is defined by <math>a_0 = 1</math>, <math>a_1 = -4</math>, and <math>a_{n+2} = -4a_{n+1} - 7a_n, \text{for } n \geq 0.</math> Find the number of positive integer divisors of <math>a_{50}^2 - a_{49}a_{51}</math>.

Revision as of 13:47, 27 April 2024

Problem 1

Let $n$ be a positive integer such that $1 \leq n \leq 1000$. Let $M_n$ be the number of integers in the set

$X_n = \left\{\sqrt{4n + 1}, \sqrt{4n + 2}, \ldots, \sqrt{4n + 1000}\right\}$. Let $a = \max\{M_n : 1 \leq n \leq 1000\}$, and $b = \min\{M_n : 1 \leq n \leq 1000\}$.

Find $a - b$.

Solution

All of the sets from $1 \leq n \leq 1000$ will have 1000 elements the most the number of square numbers will be in $X_1$ and least number of perfect squares in $X_{1000}$. So $a=M_1$ and $b=M_{1000}$. This is because the gaps between square number is less when n is greater then when n is less. By checking for $X_1$ there would be squares from {$3^2, 4^2,..., 31^2$}, a total of 29 numbers while in $X_{1000}$ there would be squares from {$64^2, 65^2,..., 70^2$} a total of 7 numbers, so $a=29$ and $b=7$, giving us $\boxed{a-b=\textbf{22}}$

~ Lakshya Pamecha

Problem 2

Find the number of elements in the set

\[\lbrace(a.b)\in N: 2 \leq a,b \leq2023,\:\: \log_{a}{b}+6\log_{b}{a}=5\rbrace\]

Problem 3

Let α and β be positive integers such that\[\frac{16}{37}<\frac{\alpha}{\beta}<\frac{7}{16}\]Find the smallest possible value of β .

Problem 4

Let $x, y$ be positive integers such that\[x^{4}=(x-1)(y^{3}-23)-1\] Find the maximum possible value of $x + y$.

Problem 5

Problem 6

Problem 7

Unconventional dice are to be designed such that the six faces are marked with numbers from $1$ to $6$ with $1$ and $2$ appearing on opposite faces. Further, each face is colored either red or yellow with opposite faces always of the same color. Two dice are considered to have the same design if one of them can be rotated to obtain a dice that has the same numbers and colors on the corresponding faces as the other one. Find the number of distinct dice that can be designed.

Solution

We can fix 1 and 2 anywhere now we can permute the remaining faces like in a necklace (not a circle because anticlockwise and clockwise would be same), so the number of ways to permute them around the remaining 4 faces would be $\frac{3!}{2}$ and number of ways to colour them $2\cdot2\cdot2$ ways. So the answer would be $\boxed{\frac{3!}{2} \cdot8=\textbf{24}}$


~Lakshya Pamecha

Problem 8

Given a 2 x 2 tile and seven dominoes (2 x 1 tile), find the number of ways of tilling a 2 x 7 rectangle using some of these tiles.

Solution

Let $a_n$ be the total number of ways of tiling a 2 x n rectangle using only dominoes. So $a_n=a_{n-1}+a_{n-2}$. Obviously $a_1=1$ and $a_2=2$ (remember right now we are not considering the 2 x 2 tile). This will give us $a_7=21$. Now we introduce the 2 x 2 tile. Number of ways of tiling the 2 x 7 board will totally depend on the position of the 2 x 2 tile.

  • Case I: If the 2 x 2 tile is on the 1st column of the 2 x 7 board then the numbers of ways for tiling the rest of the board would be $a_5=8$.
  • Case II: If the 2 x 2 tile is on the 2nd column of the 2 x 7 board then numbers of ways of tiling the rest of the board will be $a_1a_4=5$.
  • Case III: If the 2 x 2 tile is on the 3rd column of the board then the number of ways of tiling the rest of the board will be $a_2a_3=6$.

Now we will add all the cases and multiply by 2 (Why? Because the cases will repeat again). $2(8+5+6)=38$, remember this is when we consider a 2 x 2 tile, without the 2 x 2 tile there are $a_7=21$ ways. So the total number of ways of tiling a 2 x 7 board using a 2 x 2 tile and 2 x 1 dominos is $\boxed{38+21=\textbf{59}}$

~Lakshya Pamecha

Problem 9

Find the number of triples $(a, b, c)$ of positive integers such that (a) $ab$ is a prime;

(b) $bc$ is a product of two primes;

(c) $abc$ is not divisible by square of any prime and

(d) $abc\leq30$

Solution

ab is a prime means than one of ab is 1.

  • Case I: $a=1$ so b is a prime and bc is product of two pimes. Also, $bc \le 30$ and $bc\neq {4, 9, 25}$ and (b,c) are different primes. This gives {$b=2,3,5,7$} and {$c =3,5,7$}, on counting the total number of cases comes out to be 14.
  • Case II: when b=1 only 3 triplets which will satisfy the above conditions are: (2,1,15), (3,1,10), (5,1,6).

Number of triplets: $\boxed{14+3=\textbf{17}}$

Problem 10

The sequence $\{a_n\}_{n\geq0}$ is defined by $a_0 = 1$, $a_1 = -4$, and $a_{n+2} = -4a_{n+1} - 7a_n, \text{for } n \geq 0.$ Find the number of positive integer divisors of $a_{50}^2 - a_{49}a_{51}$.

Problem 11

Problem 12

Problem 13

The ex-radii of a triangle are 10 1/ 2, 12 and 14. If the sides of the triangle are the roots of the cubic $x^3-px^2+qx-r=0$ , where $p, q, r$ are integers, find the integer nearest to $\sqrt{p+q+r}$

==Problem 10

Problem 15

Problem 16

Problem 17

Consider the set\[S = \{ (a, b, c, d, e) : 0 < a < b < c < d < e < 100 \}\]where $a, b, c, d, e$ are integers. If $D$ is the average value of the fourth element of such a tuple in the set, taken over all the elements of S, find the largest integer less than or equal to D.

Problem 18

Let $P$ be a convex polygon with $50$ vertices. A set $F$ of diagonals of $P$ is said to be minimally friendly if any diagonal $d$ ∈ $F$ intersects at most one other diagonal in $F$ at a point interior to $P$ . Find the largest possible number of elements in a minimally friendly set $F$.

Problem 19

For $n$ ∈ $N$ , let $P(n)$ denote the product of the digits in $n$ and $S(n)$ denote the sum of the digits in $n$ . Consider the set $A=\{n \in N : P(n) \text{is non-zero, square free and } S(n) \text{ is a proper divisor of } P(n) \}$ . Find the maximum possible number of digits of the numbers in $A$ .

Problem 20

For any finite non empty set $X$ of integers, let $\text{max} (X)$ denote the largest element of $X$ and $|X|$ denote the number of elements in $X$ . If $N$ is the number of ordered pairs $(A, B)$ of finite non-empty sets of positive integers, such that $\text{max} (A)$ × $|B|=12$; and $|A|$ × $\text{max} (B)=11$ and $N$ can be written as $100a + b$ where $a, b$ are positive integers less than $100$, find $a + b$ .

Problem 21

For $n$ ∈ $N$ , consider non-negative integer-valued functions $f$ on ${1, 2, . . . , n}$ satisfying $f(i) \le f(j)$ for $i > j$ and Pn i=1 (i + f(i))=2023 . Choose $n$ such that Pn i=1 f(i) is the least. How many such functions exist in that case?

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

In the land of Binary, the unit of currency is called Ben and currency notes are available in denominations 1, 2, 2 2 , 2 3 , . . . Bens. The rules of the Government of Binary stipulate that one can not use more than two notes of any one denomination in any transaction. For example, one can give a change for 2 Bens in two ways: 2 one Ben notes or 1 two Ben note. For 5 Ben one can give 1 one Ben note and 1 four Ben note or 1 one Ben note and 2 two Ben notes. Using 5 one Ben notes or 3 one Ben notes and 1 two Ben notes for a 5 Ben transaction is prohibited. Find the number of ways in which one can give change for 100 Bens, following the rules of the Government.

Problem 27

Problem 28

On each side of an equilateral triangle with side length n units, where n is an integer, $1 \le n \le 100$ , consider $n-1$ points that divide the side into n equal segments. Through these points, draw lines parallel to the sides of the triangle, obtaining a net of equilateral triangles of side length one unit. On each of the vertices of these small triangles, place a coin head up. Two coins are said to be adjacent if the distance between them is 1 unit. A move consists of flipping over any three mutually adjacent coins. Find the number of values of n for which it is possible to turn all coins tail up after a finite number of moves.

Problem 29

A positive integer $n > 1$ is called $beautiful$ if $n$ can be written in one and only one way as $n = a_1 + a_2 +... + a_k = a_1 a_2 ... a_k$ for some positive integers $a_1, a_2, . . . , a_k$ , where $k > 1$ and $a_1 \geq a_2 \geq ... \geq a_k$ . (For example 6 is beautiful since 6 = 3 · 2 · 1 = 3 + 2 + 1 , and this is unique. But 8 is not beautiful since 8 = 4 + 2 + 1 + 1 = 4 · 2 · 1 · 1 as well as 8 = 2 + 2 + 2 + 1 + 1 = 2 · 2 · 2 · 1 · 1 , so uniqueness is lost.) Find the largest beautiful number less than 100.

Problem 30

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