Olimpiada de Mayo

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 $13$ years the year before, and level two is for students that have not reached the age of $15$ years the year before. The test is taken by $12$ latinamerican countries.

Member Countries

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

Past Tests/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 $388$ and end in $388$ ? 2nd level, 3rd olympiad test

Solutions

Solution to Problem #1

Let the two numbers be $a$ and $b$ such that $a > b$ .

We know that $a - b = 7$ and $ab\equiv 0\pmod{5}$ , since only one of the two numbers can be a multiple of $5$ we can start with cases, one in which $a$ is a multiple of $5$ and the other where $b$ is a multiple of $5$ .

Case 1: ( $a$ is a multiple of $5$ )

The lowest value for  that satisfies the conditions is .

The highest value for  that satisfies the conditions is .

So there are a total of $20 - 2 + 1 = 19$ cases that work.

Case 2: ( $b$ is a multiple of $5$ )

The lowest value for  that satisfies the conditions is .

The highest value for  that satisfies the conditions is .

So there are a total of $18$ cases that work.

In total there are $19 + 18 = \boxed{37}$ cases that work.

Solution to Problem #2

First we look for the Least common multiple of $388$ that ends in $3$ zeros so that when added to a number that ends in $388$ it will still be a multiple of $388$ and end in $388$ , and this number is $97000=97\cdot 1000=388\cdot 25$ .