# Difference between revisions of "Olimpiada de Mayo"

Mathgeek2006 (talk | contribs) m (→Solution to Problem #2) |
(→Member Countries) |
||

Line 6: | Line 6: | ||

* Bolivia | * Bolivia | ||

* Colombia | * Colombia | ||

+ | * Costa Rica | ||

* Ecuador | * Ecuador | ||

* El Salvador | * El Salvador | ||

+ | * España | ||

* México | * México | ||

* Panamá | * Panamá | ||

* Paraguay | * Paraguay | ||

* Perú | * Perú | ||

+ | * Portugal | ||

* Puerto Rico | * Puerto Rico | ||

* Venezuela | * Venezuela | ||

+ | |||

==Past Tests/Results== | ==Past Tests/Results== | ||

* The [http://www.oma.org.ar/enunciados/may3.htm 1997 test] | * The [http://www.oma.org.ar/enunciados/may3.htm 1997 test] |

## Latest revision as of 14:52, 17 August 2020

Olimpiada de Mayo(May Olympiad) is an annual test that has two levels, level one is for students that have not reached the age of years the year before, and level two is for students that have not reached the age of years the year before. The test is taken by latinamerican countries.

## Contents

## Member Countries

- Argentina
- Brasil
- Bolivia
- Colombia
- Costa Rica
- Ecuador
- El Salvador
- España
- México
- Panamá
- Paraguay
- Perú
- Portugal
- Puerto Rico
- Venezuela

## Past Tests/Results

- The 1997 test
- The 1998 test, 1998 test results
- The 1999 test
- The 2000 test
- The 2001 test
- The 2002 test
- The 2003 test
- The 2004 test
- The 2005 test
- The 2006 test, 2006 test results
- The 2007 test, 2007 test results
- The 2008 test, 2008 test results
- The 2009 test, 2009 test results

## Sample Problems

Some problems from previous Olimpiada de Mayo tests.

### Problem 1

We choose 2 numbers in between 1 and 100, inclusive, such that their difference is 7 and their product is a multiple of 5. How many ways can we do this? *1st level, 5th olympiad test*

### Problem 2

How many 7-digit numbers are multiples of and end in ? *2nd level, 3rd olympiad test*

## Solutions

### Solution to Problem #1

Let the two numbers be and such that .

We know that and , since only one of the two numbers can be a multiple of we can start with cases, one in which is a multiple of and the other where is a multiple of .

Case 1: ( is a multiple of )

The lowest value for that satisfies the conditions is .

The highest value for that satisfies the conditions is .

So there are a total of cases that work.

Case 2: ( is a multiple of )

The lowest value for that satisfies the conditions is .

The highest value for that satisfies the conditions is .

So there are a total of cases that work.

In total there are cases that work.

### Solution to Problem #2

First we look for the Least common multiple of that ends in zeros so that when added to a number that ends in it will still be a multiple of and end in , and this number is .

So any number in the form of ends in and is a multiple of .

We want to find

Since we are only dealing with the thousands we don't need to worry about the :

From this we see that there are solutions.