Title: There is a sports competition containing M items, athletes a,b,c participation, in each project, first, second, the third place respectively p1,p2,p3 points, wherein P1,P2,P3 is a positive integer and P1>P2>P3. Finally a score of 22 points, B and C are 9 points, B in the hundred-meter race to achieve the first. Ask for the value of M, and asked in the high jump who got second place.
Analysis:
1, according to test instructions, can get the equation:
M (P1+P2+P3) =22+9+9=40①
P1+p2+p3≥1+2+3=6②
∴6m≤m (P1+P2+P3) = 40, so m≤6.
2, in addition by test instructions know at least hundred meters and high jump two items, thereby m≥2.
3, M need to be 40 of the approximate, in accordance with 2 m desirable: 2,4,5.
4, consider 3 kinds of cases, respectively:
A, m=2, then only high jump and hundred meters, and b hundred meters first, but the total score of only 9 points, it must have: 9≥P1+P3, so a can not get 22 points;
B, m=4, from B know: 9≥p1+3p3, and p3≥1, so p1≤6, if p1≤5, then four items up to 20 points, a can not get 22 points, so p1=6.
∵4 (P1+P2+P3) =40,∴p2+p3=4.
There are: P2=3,p3=1,a up to three first, a second, a total score 3x6+3=21<22, contradictions.
C, m=5, at this time by 5 (P1+P2+P3) = 40, Get:
P1+p2+p3=8. If p3≥2, then:
P1+p2+p3≥4+3+2=9, contradiction, so p3=1.
Also because P1 must be greater than or equal to 5, otherwise, a five times the highest can only get 20 points, and the problem set contradiction, so p1≥5.
In addition if p1≥6, then p2+p3≤2, this also with the problem set contradiction, ∴p1=5,p2+p3=3, namely p2=2,p3=1.
A=22=4x5+2, so a got four first, a second, and because B hundred meters got first, so hundred meters second must be a.
B=9=5+4x1, then B got a first, four third;
C=9=4x2+1, C got four second, one third.
C gets second place in the high jump
Answer: M=5, C takes second place in the high jump.
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