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Difference between revisions of "Olimpiada de Mayo"
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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. Twelve latin-american countries compete in the test, these countries are:
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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 <math>13</math> years the year before, and level two is for students that have not reached the age of <math>15</math> years the year before. The test is taken by <math>12</math> latinamerican countries.
  
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==Member Countries==
 
* Argentina
 
* Argentina
 
* Brasil
 
* Brasil
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* Puerto Rico
 
* Puerto Rico
 
* Venezuela
 
* Venezuela
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==Past Tests/Results==
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* The [http://www.oma.org.ar/enunciados/may3.htm 1997 test], [http://www.oma.org.ar/internacional/resultados-may3.htm 1997 test results]
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* The [http://www.oma.org.ar/enunciados/may4.htm 1998 test], [http://www.oma.org.ar/internacional/resultados-may4.htm 1998 test results]
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* The [http://www.oma.org.ar/enunciados/may5.htm 1999 test], [http://www.oma.org.ar/internacional/resultados-may5.htm 1999 test results]
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* The [http://www.oma.org.ar/enunciados/may6.htm 2000 test], [http://www.oma.org.ar/internacional/resultados-may6.htm 2000 test results]
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* The [http://www.oma.org.ar/enunciados/may7.htm 2001 test], [http://www.oma.org.ar/internacional/resultados-may7.htm 2001 test results]
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* The [http://www.oma.org.ar/enunciados/may8.htm 2002 test], [http://www.oma.org.ar/internacional/resultados-may8.htm 2002 test results]
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* The [http://www.oma.org.ar/enunciados/may9.htm 2003 test], [http://www.oma.org.ar/internacional/resultados-may9.htm 2003 test results]
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* The [http://www.oma.org.ar/enunciados/may10.htm 2004 test], [http://www.oma.org.ar/internacional/resultados-may10.htm 2004 test results]
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* The [http://www.oma.org.ar/enunciados/may11.htm 2005 test], [http://www.oma.org.ar/internacional/resultados-may11.htm 2005 test results]
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* The [http://www.oma.org.ar/enunciados/may12.htm 2006 test], [http://www.oma.org.ar/internacional/resultados-may12.htm 2006 test results]
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* The [http://www.oma.org.ar/enunciados/may13.htm 2007 test], [http://www.oma.org.ar/internacional/resultados-may13.htm 2007 test results]
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* The [http://www.oma.org.ar/enunciados/may14.htm 2008 test], [http://www.oma.org.ar/internacional/resultados-may14.htm 2008 test results]
 
==Problems==
 
==Problems==
 
Some problems from previous Olimpiada de Mayo tests.
 
Some problems from previous Olimpiada de Mayo tests.
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[http://www.oma.org.ar/internacional/may.htm Olimpiada de Mayo home]
 
[http://www.oma.org.ar/internacional/may.htm Olimpiada de Mayo home]
  
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[http://www.oma.org.ar/enunciados/index.htm#may practice tests]
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[http://www.oma.org.ar/enunciados/index.htm#may Practice tests]

Revision as of 00:10, 18 May 2009

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

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. Unknown error_msg)

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

Reference

Olimpiada de Mayo home

Practice tests

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