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 \(n\).

## Comments

Sort by:

TopNewestI like question 2! – Calvin Lin Staff · 2 years, 5 months ago

Log in to reply

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, 5 months ago

Log in to reply

– Anusmita Mukherjee · 2 years, 4 months ago

can u elaborate the answer for the first question , please?Log in to reply

– Jordi Bosch · 2 years, 5 months ago

Under which condition does \(7|2n+1\)?Log in to reply

– Siddhartha Srivastava · 2 years, 5 months ago

When \( 2n + 1 \equiv 0 \pmod7 \) or \( 2n \equiv 6 \pmod7 \) or \( n \equiv 3 \pmod7 \)Log in to reply

– Jordi Bosch · 2 years, 5 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 \((n^2+n+1)\).Log in to reply

picture

Log in to reply

– Jordi Bosch · 2 years, 5 months ago

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

I want learn these sums with enthusiasm. – Anuvind Shrivastava · 2 years, 5 months ago

Log in to reply

for 2 i got the solutions: (x.y.z)=( -2,-3,6) – Akash Deep · 2 years, 5 months ago

Log in to reply

– Calvin Lin Staff · 2 years, 5 months ago

Are there any other solutions?Log in to reply

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

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

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

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

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

\((6,-3,-2)\) – Krishna Sharma · 2 years, 5 months ago

Log in to reply

– Akash Deep · 2 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 resultLog in to reply

– Krishna Sharma · 2 years, 5 months ago

Correct :)Log in to reply