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
3 years, 4 months ago

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I like question 2!

Staff - 3 years, 4 months ago

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 $$7|2n+1$$

- 3 years, 4 months ago

- 3 years, 3 months ago

Under which condition does $$7|2n+1$$?

- 3 years, 4 months ago

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

- 3 years, 4 months ago

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 $$(n^2+n+1)$$.

- 3 years, 4 months ago

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?

- 3 years, 4 months ago

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

- 3 years, 4 months ago

I want learn these sums with enthusiasm.

- 3 years, 4 months ago

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

- 3 years, 4 months ago

Are there any other solutions?

Staff - 3 years, 4 months ago

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

$$(-2,-3,6)$$

$$(-2, 6, -3)$$

$$(-3,-2,6)$$

$$(-3,6,-2)$$

$$(6,-2,-3)$$

$$(6,-3,-2)$$

- 3 years, 4 months ago

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

- 3 years, 3 months ago

Correct :)

- 3 years, 4 months ago