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:

${5\choose2}\cdot{6\choose3-2}\cdot{11\choose4-3}\cdot{23\choose5-4}= {5\choose2}\cdot{6\choose1}\cdot{11\choose1}\cdot{23\choose1}={5\cdot4\over2}\cdot6\cdot11\cdot23=15180.$

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

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.

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