Problem K. 411. (February 2014)
K. 411. The power x^{5} cannot only be calculated by doing the multiplication x^{.}x^{.}x^{.}x^{.}x, but also with fewer multiplications, by calculating partial results (e.g. y=x^{.}x) and obtaining the final answer by multiplying those: y^{.}y^{.}x, which only makes three multiplications altogether. Express the power x^{23} with less than 10 multiplications to carry out.
(6 pont)
Deadline expired on 10 March 2014.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. Egy lehetséges út 7 szorzással: x^{.}x=x^{2}, x^{2}^{.}x^{2}=x^{4}, x^{4}^{.}x^{4}=x^{8}, x^{8}^{.}x^{8}=x^{16}, x^{16}^{.}x^{4}^{.}x^{2}^{.}x=x^{23}.
Egy lehetséges út 6 szorzással: x^{.}x=x^{2}, x^{2}^{.}x=x^{3}, x^{2}^{.}x^{3}=x^{5}, x^{5}^{.}x^{5}=x^{10}, x^{10}^{.}x^{3}=x^{13}, x^{10}^{.}x^{13}=x^{23}.
Statistics:
161 students sent a solution.  
6 points:  109 students. 
5 points:  16 students. 
4 points:  26 students. 
3 points:  2 students. 
2 points:  3 students. 
1 point:  1 student. 
0 point:  2 students. 
Unfair, not evaluated:  1 solution. 
Unfair, not evaluated:  1 solution. 
