Problem C. 829. (November 2005)
C. 829. The five numbers drawn in the lottery of 10 September, 2005 were as follows: 4, 16, 22, 48, 88. All the five numbers are even, exactly four of them are divisible by four, three are divisible by 8 and two by 16. In how many different ways is it possible to select five numbers out of the whole numbers 1 to 90?
(5 pont)
Deadline expired on December 15, 2005.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás: 1-től 90-ig 5 db 16-tal osztható szám van; 11-5=6 db olyan, ami 8-cal osztható, de 16-tal nem; 22-11=11 olyan, ami 4-gyel osztható, de 8-cal nem; végül 45-22=23 olyan, ami páros, de nem osztható 4-gyel. A feladat feltételeinek megfelelő húzások száma tehát:
Statistics:
428 students sent a solution. 5 points: 84 students. 4 points: 154 students. 3 points: 64 students. 2 points: 43 students. 1 point: 40 students. 0 point: 12 students. Unfair, not evaluated: 31 solutionss.
Problems in Mathematics of KöMaL, November 2005