What is the smallest number of additional weeks that will allow them to have the same score (see below)? Include an explanation and reasons please! Aisha, Ben and Cheng entered a weekly school competition. In each week they participated, the three of them came first, second and third in some order. The score for 1st place was 8, for 2nd place it was 2 and for 3rd place it was 1. At one stage, the total score for Aisha was 12, for Ben it was 34 and for Cheng 9. After a few weeks, Aisha, Ben and Cheng had the same score. What is the smallest number of additional weeks that will allow them to have the same score? Please include an explanation and reasons! Thank you!

Expert Answers

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Justification:

Cheng is 25 points behind Ben.

Aisha is 22 points behind Ben.

Together, they are 47 points behind Ben.

Together they can win at most 10 (8+2) points and ben will win at least 1 point.

At each game they can win 9 points more than Ben.

They are 47 points behind wich mean they need to play at least 6 games to catch up with him (round up 47/9)

The number of additional games is at least 6.

 

 

After each game 11 points are added to the total of their scores.

If the players have the same score, the total of their scores is a multiple of 3.

Let n be the number of additional games

12+9+34+11n is a multiple of 3

iff 12+9+33+1+11n is a multiple of 3

iff 1+11n is a multiple of 3.

iff n=1+3k for any `k in NN`

 

Therefore, the number of games will be 1, or 4, or 7, or 10....

In conclusion, The number of additional games  is at least 6, in the form of 1+3k

We have a solution with 7 games

Therefore the munimum number of additional games is 7.

 

 

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A solution in 7 additional weeks:

Aisha 8 8 8 2 2 2 2

Ben   2 2 2 1 1 1 1

Cheng1 1 1 8 8 8 8

 

 

Approved by eNotes Editorial Team

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