Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Problem K. 329. (February 2012)

K. 329. Given that x is a positive real number such that , determine the value of without finding the value of x.

(6 pont)

Deadline expired on March 12, 2012.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Használjuk $\displaystyle \left( x + \frac {1}{x} \right)$ hatványait: $\displaystyle \left( x + \frac {1}{x} \right)^2=x^2 + \frac {1}{x^2} + 2 =7+2=9$, azaz ($\displaystyle x$ pozitív!) $\displaystyle x + \frac {1}{x}=3$. $\displaystyle \left( x + \frac {1}{x}\right)^5=x^5+5x^3+10x+\frac{10}x +\frac{5}{x^3}+\frac{1}{x^5}$, illetve $\displaystyle \left( x + \frac {1}{x} \right)^3=x^3+3x+\frac{3}{x} +\frac{1}{x^3}$. A harmadik hatványban felhasználva korábbi eredményünket $\displaystyle 3^3=x^3+\frac{1}{x^3}+3\cdot 3$, ahonnan $\displaystyle x^3+\frac{1}{x^3}=27-9=18$. Ezt felhasználva nézzük az ötödik hatványt: $\displaystyle 3^5=x^5+\frac{1}{x^5}+5\cdot 18 +10 \cdot 3$, azaz $\displaystyle x^5+\frac{1}{x^5}=243-90-30=123$.

Statistics:

 118 students sent a solution. 6 points: 67 students. 5 points: 27 students. 4 points: 1 student. 3 points: 2 students. 2 points: 2 students. 1 point: 5 students. 0 point: 13 students. Unfair, not evaluated: 1 solutions.

Problems in Mathematics of KöMaL, February 2012