View source for 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 <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.

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

==Past Tests/Results==
* The [http://www.oma.org.ar/enunciados/may3.htm 1997 test]
* The [http://www.oma.org.ar/enunciados/may4.htm 1998 test], [http://www.oma.org.ar/internacional/resultados_generales-may4.htm 1998 test results]
* The [http://www.oma.org.ar/enunciados/may5.htm 1999 test]
* The [http://www.oma.org.ar/enunciados/may6.htm 2000 test]
* The [http://www.oma.org.ar/enunciados/may7.htm 2001 test]
* The [http://www.oma.org.ar/enunciados/may8.htm 2002 test]
* The [http://www.oma.org.ar/enunciados/may9.htm 2003 test]
* The [http://www.oma.org.ar/enunciados/may10.htm 2004 test]
* The [http://www.oma.org.ar/enunciados/may11.htm 2005 test]
* The [http://www.oma.org.ar/enunciados/may12.htm 2006 test], [http://www.oma.org.ar/internacional/resultados_generales-may12.htm 2006 test results]
* The [http://www.oma.org.ar/enunciados/may13.htm 2007 test], [http://www.oma.org.ar/internacional/resultados_generales-may13.htm 2007 test results]
* The [http://www.oma.org.ar/enunciados/may14.htm 2008 test], [http://www.oma.org.ar/internacional/resultados_generales-may14.htm 2008 test results]
* The [http://www.oma.org.ar/enunciados/may15.htm 2009 test], [http://www.oma.org.ar/internacional/resultados_generales-may15.htm 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 <math>388</math> and end in <math>388</math>? ''2nd level, 3rd olympiad test''

==Solutions==

===Solution to Problem #1===

Let the two numbers be <math>a</math> and <math>b</math> such that <math>a > b</math>.

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

Case 1: (<math>a</math> is a multiple of <math>5</math>)

 The lowest value for <math>a</math> that satisfies the conditions is <math>10 = 5(2)</math>.

 The highest value for <math>a</math> that satisfies the conditions is <math>100 = 5(20)</math>.

So there are a total of <math>20 - 2 + 1 = 19</math> cases that work.

Case 2: (<math>b</math> is a multiple of <math>5</math>)

 The lowest value for <math>b</math> that satisfies the conditions is <math>5=5(1)</math>.

 The highest value for <math>b</math> that satisfies the conditions is <math>90 = 5(18)</math>.

So there are a total of <math>18</math> cases that work.


------------------------------------------------

In total there are <math>19 + 18 = \boxed{37}</math> cases that work.

===Solution to Problem #2===

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

So any number in the form of <math>3880388 + 97000k</math> ends in <math>388</math> and is a multiple of <math>388</math>.

We want to find <math>1000000 < 3880388 + 97000k < 10000000</math> 

Since we are only dealing with the thousands we don't need to worry about the <math>388</math>:

<cmath>\begin{eqnarray*} 1000 & < & 3880 + 97k < 100000 \\
1000 & < & 97(40 + k) < 10000 \\
10 & < & 40 + k < 103 \\
- 30 & < & k < 63 \end{eqnarray*}</cmath>

From this we see that there are <math>\boxed{92}</math> solutions.

==Reference==

[http://www.oma.org.ar/internacional/may.htm Olimpiada de Mayo home]

[http://www.oma.org.ar/enunciados/index.htm#may Practice tests]