Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem K. 411. (February 2014)

K. 411. The power x5 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 x23 with less than 10 multiplications to carry out.

(6 pont)

Deadline expired on March 10, 2014.

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

Megoldás. Egy lehetséges út 7 szorzással: x.x=x2, x2.x2=x4, x4.x4=x8, x8.x8=x16, x16.x4.x2.x=x23.

Egy lehetséges út 6 szorzással: x.x=x2, x2.x=x3, x2.x3=x5, x5.x5=x10, x10.x3=x13, x10.x13=x23.

### Statistics:

 160 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 solutions.

Problems in Mathematics of KöMaL, February 2014