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Me and my friend came across this maths puzzle and are stumped to solve it. "You're in a race with 1/5 of the racers ahead of you and 5/6 of them behind you. How are people are in the race?" The only additional information was it was a round running track! Marty
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Hi Marty, I want to change your problem slightly, solve the modified problem and then come back to your problem. here is my modified problem. "You're in a race with 1/7 of the racers ahead of you and 5/6 of them behind you. How are people are in the race?" Since 1/7 of the racers are ahead of you the number of people in the race must be divisible by 7. Also, since 5/6 of the racers are behind of you the number of people in the race must be divisible by 6. The smallest number that is divisible by both 6 and 7 is 42, so does 42 work? If there are 42 people in the race then
This is everyone in the race except you so 42 is the answer to my problem. (I choose 42 as the smallest number divisible by 6 and 7. Why not 2 Now let's try your problem. "You're in a race with 1/5 of the racers ahead of you and 5/6 of them behind you. How are people are in the race?" Using the same argument, the number of people in the race is divisible by both 5 and 6. The smallest number divisible by both 5 and 6 is 30. If there are 30 people in the race then
But there are only 30 people on the race so 30 doesn't work. The next smallest number that is divisible by both 5 and 6 is 2 1/5 of the people are ahead of you and 5/6 of the people are behind you and
But 31/30 is larger than 1, so if you take the number of people ahead of you plus the number of people behind you the result is more people than are in the race. The situation described in the problem is impossible. Penny
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42 = 6, so 6 people are ahead of you |
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