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Regonial Spanish Olympiad

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 \[\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx} = \frac{1}{6^{2}}\]\[\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy} = \frac{7^{2}}{6^{2}}\]\[\frac{1}{(xy)^{2}}+\frac{1}{(yz)^{2}}+ \frac{1}{(zx)^{2}} = \frac{7^{2}}{36^{2}}\]

3 For every positive integer \(n \geq 1\) we denote \(a_{n} = n^4 + n^2 + 1\). Find the greatest common divisor of \( a_{n}\) and \(a_{n+1}\) in function of \(n\).

Note by Jordi Bosch
2 years, 10 months ago

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I like question 2!

Calvin Lin Staff - 2 years, 10 months ago

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For 1. I got \( 2*3^n \),

for 2. I got \( (x,y,z) = (-2,-3,6) \) and permutations and

for 3. I got \( n^2 + n + 1 \) if \( 7 \not | 2n+1 \) and \( 7(n^2 + n + 1) \) if \( 7|2n+1 \)

Siddhartha Srivastava - 2 years, 10 months ago

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can u elaborate the answer for the first question , please?

Anusmita Mukherjee - 2 years, 9 months ago

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Under which condition does \(7|2n+1\)?

Jordi Bosch - 2 years, 10 months ago

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When \( 2n + 1 \equiv 0 \pmod7 \) or \( 2n \equiv 6 \pmod7 \) or \( n \equiv 3 \pmod7 \)

Siddhartha Srivastava - 2 years, 10 months ago

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@Siddhartha Srivastava Nicely done! that is if \(n \equiv 3 \pmod{7}\) then the greatest common divisor of \(a_{n}\) and \(a_{n+1}\) is \(7(n^2+n+1)\), else the gcd between \(a_{n}\) and \(a_{n+1}\) is \((n^2+n+1)\).

Jordi Bosch - 2 years, 10 months ago

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picture

picture

For 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?

Adarsh Kumar - 2 years, 10 months ago

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The problem quickly follows if you consider that a row is already painted and you study the possibilities for next row.

Jordi Bosch - 2 years, 10 months ago

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I want learn these sums with enthusiasm.

Anuvind Shrivastava - 2 years, 10 months ago

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for 2 i got the solutions: (x.y.z)=( -2,-3,6)

Akash Deep - 2 years, 10 months ago

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Are there any other solutions?

Calvin Lin Staff - 2 years, 10 months ago

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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

\((-2,-3,6)\)

\((-2, 6, -3)\)

\((-3,-2,6)\)

\((-3,6,-2)\)

\((6,-2,-3)\)

\((6,-3,-2)\)

Krishna Sharma - 2 years, 10 months ago

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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

Akash Deep - 2 years, 9 months ago

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Correct :)

Krishna Sharma - 2 years, 10 months ago

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