These are some of the Regional Spanish Mathematical Olympiad
1 We have a rectangle of 2∗n squares, in n columns of 2 squares each. We have 3 colors to paint the squares, and we want that every two squares that share a side don't have the same color. How many diferent ways are there?
2 Find the real solutions of xy1+yz1+zx1=621yzx+xzy+xyz=6272(xy)21+(yz)21+(zx)21=36272
3 For every positive integer n≥1 we denote an=n4+n2+1. Find the greatest common divisor of an and an+1 in function of n.
#Algebra
#NumberTheory
#OlympiadMath
#Catalan
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
I like question 2!
pictureFor the square ABCD we have three choices.For the square BCFE we have two choices.Similarly,for the square GDCH we have two choices.Now,for the square HCIF we have two choices or one choice depending on the fact whether the squares,GDCH and BCFE have the same colour or different colours.WHAT TO DO NOW?
Log in to reply
The problem quickly follows if you consider that a row is already painted and you study the possibilities for next row.
For 1. I got 2∗3n,
for 2. I got (x,y,z)=(−2,−3,6) and permutations and
for 3. I got n2+n+1 if 7∣2n+1 and 7(n2+n+1) if 7∣2n+1
Log in to reply
Under which condition does 7∣2n+1?
Log in to reply
When 2n+1≡0(mod7) or 2n≡6(mod7) or n≡3(mod7)
Log in to reply
n≡3(mod7) then the greatest common divisor of an and an+1 is 7(n2+n+1), else the gcd between an and an+1 is (n2+n+1).
Nicely done! that is ifcan u elaborate the answer for the first question , please?
for 2 i got the solutions: (x.y.z)=( -2,-3,6)
Log in to reply
Correct :)
Are there any other solutions?
Log in to reply
We can make a cubic polynomial with x,y,z as root, we get can get 6 pairs of x,y,z(3 roots all real) by interchanging roots
(−2,−3,6)
(−2,6,−3)
(−3,−2,6)
(−3,6,−2)
(6,−2,−3)
(6,−3,−2)
no as @krishna sharma says we can obtain the values for x+y+z, xy+yz+xz and xyz and then declare a polynomial with roots x,y,z and place the coefficient values which will be x+y+z, xy+yz+xz and xyz and then use factor theorem to get the result
I want learn these sums with enthusiasm.