Olimpiada de Mayo

Revision as of 23:59, 9 June 2010 by Tonypr (talk | contribs) (Problems)

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
  • Ecuador
  • El Salvador
  • México
  • Panamá
  • Paraguay
  • Perú
  • 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 $a$ 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 $388$:

$\begin{eqnarray*} 1000 & < & 3880 + 97k < 100000 \\ 1000 & < & 97(40 + k) < 10000 \\ 10 & < & 40 + k < 103 \\ - 30 & < & k < 63 \end{eqnarray*}$ (Error compiling LaTeX. ! Missing \endgroup inserted.)

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

Reference

Olimpiada de Mayo home

Practice tests

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