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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 points)

This problem is for grade 9 students only.

Deadline expired on 10 March 2014.

Google Translation (Sorry, the solution is published in Hungarian only.)

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 on problem K. 411.
 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.

• Problems in Mathematics of KöMaL, February 2014

•  Támogatóink: Morgan Stanley