Problem B. 4304. (November 2010)

B. 4304. Is there a positive integer k, such that $\big(\dots\big((4\underbrace{!)!\big)!\dots\big)!}_{k} > \big(\dots\big((3\underbrace{!)!\big)!\dots\big)!}_{k+1}$ ?

(3 pont)

Deadline expired on December 10, 2010.

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

Megoldás. Nincsen ilyen $\displaystyle k$ szám. Legyen $\displaystyle a_k=(\dots((3\underbrace{!)!)!\dots)!}_{k}$, $\displaystyle b_k=(\dots((4\underbrace{!)!)!\dots)!}_{k}$. A $\displaystyle k=1$ esetben $\displaystyle a_2=(3!)!=6!>4!=b_1$. Ha pedig valamely $\displaystyle k$ pozitív egészre $\displaystyle a_{k+1}>b_k$ pozitív egészek, akkor $\displaystyle a_{k+2}=a_{k+1}!>b_k!=b_{k+1}$ is teljesül. Így a teljes indukció elve szerint minden $\displaystyle k$ pozitív egészre

$\displaystyle (\dots((3\underbrace{!)!)!\dots)!}_{k+1}> (\dots((4\underbrace{!)!)!\dots)!}_{k}\ .$

Statistics:

209 students sent a solution.

3 points: 176 students.

2 points: 25 students.

0 point: 1 student.

Unfair, not evaluated: 7 solutionss.

Problems in Mathematics of KöMaL, November 2010

209 students sent a solution.
3 points:	176 students.
2 points:	25 students.
0 point:	1 student.
Unfair, not evaluated:	7 solutionss.

Already signed up?
New to KöMaL?