Waste less time on Facebook — follow Brilliant.
×

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, 7 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

I like question 2! Calvin Lin Staff · 2 years, 7 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, 7 months ago

Log in to reply

@Siddhartha Srivastava can u elaborate the answer for the first question , please? Anusmita Mukherjee · 2 years, 6 months ago

Log in to reply

@Siddhartha Srivastava Under which condition does \(7|2n+1\)? Jordi Bosch · 2 years, 7 months ago

Log in to reply

@Jordi Bosch When \( 2n + 1 \equiv 0 \pmod7 \) or \( 2n \equiv 6 \pmod7 \) or \( n \equiv 3 \pmod7 \) Siddhartha Srivastava · 2 years, 7 months ago

Log in to reply

@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, 7 months ago

Log in to reply

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, 7 months ago

Log in to reply

@Adarsh Kumar 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, 7 months ago

Log in to reply

I want learn these sums with enthusiasm. Anuvind Shrivastava · 2 years, 7 months ago

Log in to reply

for 2 i got the solutions: (x.y.z)=( -2,-3,6) Akash Deep · 2 years, 7 months ago

Log in to reply

@Akash Deep Are there any other solutions? Calvin Lin Staff · 2 years, 7 months ago

Log in to reply

@Calvin Lin 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, 7 months ago

Log in to reply

@Calvin Lin 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, 6 months ago

Log in to reply

@Akash Deep Correct :) Krishna Sharma · 2 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...