# Difference between revisions of "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.

• Argentina
• Brasil
• Bolivia
• Colombia
• Ecuador
• El Salvador
• México
• Panamá
• Paraguay
• Perú
• Puerto Rico
• Venezuela

## 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 $a$ that satisfies the conditions is $10 = 5(2)$.

The highest value for $a$ that satisfies the conditions is $100 = 5(20)$.


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 $b$ that satisfies the conditions is $5=5(1)$.

The highest value for $b$ that satisfies the conditions is $90 = 5(18)$.


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 $97000=97\cdot 1000=388\cdot 25$.

So any number in the form of $3880388 + 97000k$ ends in $388$ and is a multiple of $388$.

We want to find $1000000 < 3880388 + 97000k < 10000000$

Since we are only dealing with the thousands we don't need to worry about the $388$: $\begin{eqnarray*} 1000 & < & 3880 + 97k < 100000 \\ 1000 & < & 97(40 + k) < 10000 \\ 10 & < & 40 + k < 103 \\ - 30 & < & k < 63 \end{eqnarray*}$

From this we see that there are $\boxed{92}$ solutions.

Invalid username
Login to AoPS