Probability Questions with Cars

Probability theory and statistics

Probability Questions with Cars

Postby martapag » Tue Jan 22, 2013 3:05 am

Hello everyone. MY first time here Hope someone can help me. Lets say Peter has fewer than 1000 cars. With that he made 5 equal rows of cars and have one car left. He also made 4 equal rows of cars and also have one car left. He then made 9 equal rows and didnt have any cars left. He can have six possibilities. So how many cars does he have?

What is the best way to explain it.

Thank you all , Marta





martapag
 
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Re: Probability Questions with Cars

Postby Guest » Tue Jan 22, 2013 8:53 am

What you should do is express the question in terms of modula arithmetic then use the Chinese Remainder Theorem.
However we can derive the answer without explicitly using it.

Let [tex]x[/tex] be the number of cars. The second and third pieces of information say that if we divide by 5 or 4 we get a remainder of 1. So [tex]x-1[/tex] must be divisible by 4 and 5, so [tex]x-1[/tex] must be a multiple of 20. This means [tex]x = 20k+1[/tex] for some integer [tex]k[/tex]. The fourth piece of information says [tex]x[/tex] is a multiple of 9. We can always write [tex]k[/tex] as [tex]9q+r[/tex] where [tex]q[/tex] is the quotient and [tex]r[/tex] is the remainder when we divide [tex]k[/tex] by 9 (note that [tex]q[/tex] and [tex]r[/tex] are integers and furthermore [tex]r[/tex] is 0,1,2,...,or 8 ). Putting all this together tells us that
[tex]x = 20k+1 = 20(9q+r)+1 = 180q+20r+1[/tex]
is a multiple of 9. Clearly 180q is a multiple of 9 and since the entire expression is a multiple of 9 we know [tex]20r+1[/tex] is a multiple of 9. Checking the 9 possible values of [tex]r[/tex] (i.e. 0,1,...8 ) we find the only time when 20r+1 is divisible by 9 is when [tex]r = 4[/tex]. So [tex]x = 180q+81[/tex]

It is easy to check that regardless of the value of [tex]q[/tex] that [tex]180q+81[/tex] has remainder 1 when divided by 4 and 5 and remainder 0 when divided by 9.

Since we know the numbers of cars is less than 1000, the number of cars must be
81, 261, 441, 621, 801, 981

Hope this helps,

R. Baber.

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Re: Probability Questions with Cars

Postby martapag » Tue Jan 22, 2013 4:49 pm

Guest Thank you very much :)

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