Difference between revisions of "2023 IOQM Problems"
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==Solution== | ==Solution== | ||
All of the sets from <math>1 \leq n \leq 1000</math> will have 1000 elements the most the number of square numbers will be in <math>X_1</math> and least number of perfect squares in <math>X_{1000}</math>. So <math>a=M_1</math> and <math>b=M_{1000}</math>. This is because the gaps between square number is less when n is greater then when n is less. By checking for <math>X_1</math> there would be squares from {<math>3^2, 4^2,..., 31^2</math>}, a total of 29 numbers while in <math>X_{1000}</math> there would be squares from {<math>64^2, 65^2,..., 70^2</math>} a total of 7 numbers, so <math>a=29</math> and <math>b=7</math>, giving us <math>\boxed{a-b=\textbf{22}}</math> | All of the sets from <math>1 \leq n \leq 1000</math> will have 1000 elements the most the number of square numbers will be in <math>X_1</math> and least number of perfect squares in <math>X_{1000}</math>. So <math>a=M_1</math> and <math>b=M_{1000}</math>. This is because the gaps between square number is less when n is greater then when n is less. By checking for <math>X_1</math> there would be squares from {<math>3^2, 4^2,..., 31^2</math>}, a total of 29 numbers while in <math>X_{1000}</math> there would be squares from {<math>64^2, 65^2,..., 70^2</math>} a total of 7 numbers, so <math>a=29</math> and <math>b=7</math>, giving us <math>\boxed{a-b=\textbf{22}}</math> | ||
+ | |||
+ | ~ Lakshya Pamecha | ||
==Problem 2== | ==Problem 2== |
Revision as of 13:09, 27 April 2024
Contents
- 1 Problem 1
- 2 Solution
- 3 Problem 2
- 4 Problem 3
- 5 Problem 4
- 6 Problem 5
- 7 Problem 6
- 8 Problem 7
- 9 Problem 8
- 10 Solution
- 11 Problem 9
- 12 Problem 10
- 13 Problem 11
- 14 Problem 12
- 15 Problem 13
- 16 Problem 15
- 17 Problem 16
- 18 Problem 17
- 19 Problem 18
- 20 Problem 19
- 21 Problem 20
- 22 Problem 21
- 23 Problem 22
- 24 Problem 23
- 25 Problem 24
- 26 Problem 25
- 27 Problem 26
- 28 Problem 27
- 29 Problem 28
- 30 Problem 29
- 31 Problem 30
Problem 1
Let be a positive integer such that . Let be the number of integers in the set
. Let , and .
Find .
Solution
All of the sets from will have 1000 elements the most the number of square numbers will be in and least number of perfect squares in . So and . This is because the gaps between square number is less when n is greater then when n is less. By checking for there would be squares from {}, a total of 29 numbers while in there would be squares from {} a total of 7 numbers, so and , giving us
~ Lakshya Pamecha
Problem 2
Find the number of elements in the set
Problem 3
Let α and β be positive integers such thatFind the smallest possible value of β .
Problem 4
Let be positive integers such that Find the maximum possible value of .
Problem 5
Problem 6
Problem 7
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 be the total number of ways of tiling a 2 x n rectangle using only dominoes. So . Obviously and (remember right now we are not considering the 2 x 2 tile). This will give us . 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 .
- 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 .
- 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 .
Now we will add all the cases and multiply by 2 (Why? Because the cases will repeat again). , remember this is when we consider a 2 x 2 tile, without the 2 x 2 tile there are 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
~Lakshya Pamecha
Problem 9
Find the number of triples of positive integers such that (a) is a prime;
(b) is a product of two primes;
(c) is not divisible by square of any prime and
(d)
Problem 10
The sequence is defined by , , and Find the number of positive integer divisors of .
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 , where are integers, find the integer nearest to
==Problem 10
Problem 15
Problem 16
Problem 17
Consider the setwhere are integers. If 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 be a convex polygon with vertices. A set of diagonals of is said to be minimally friendly if any diagonal ∈ intersects at most one other diagonal in at a point interior to . Find the largest possible number of elements in a minimally friendly set .
Problem 19
For ∈ , let denote the product of the digits in and denote the sum of the digits in . Consider the set . Find the maximum possible number of digits of the numbers in .
Problem 20
For any finite non empty set of integers, let denote the largest element of and denote the number of elements in . If is the number of ordered pairs of finite non-empty sets of positive integers, such that × ; and × and can be written as where are positive integers less than , find .
Problem 21
For ∈ , consider non-negative integer-valued functions on satisfying for and Pn i=1 (i + f(i))=2023 . Choose 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, , consider 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 is called if can be written in one and only one way as for some positive integers , where and . (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.