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Math Help - Maximum Value

  1. #1
    Senior Member pankaj's Avatar
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    Maximum Value

    Let x_{1},x_{2},x_{3},x_{4} be the four digits of a four digit number N=x_{1}x_{2}x_{3}x_{4}.

    Find the maximum value of \frac{N}{x_{1}+x_{2}+x_{3}+x_{4}}
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  2. #2
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    Wouldn't it be 1000?

    \frac{N}{x_{1}+x_{2}+x_{3}+x_{4}} \leq N

    So, if x_{1}+x_{2}+x_{3}+x_{4} = 1, then we get N=N, which is the most that \frac{N}{x_{1}+x_{2}+x_{3}+x_{4}} can be.

    Since the number must have four digits, we can't have x_{1} = 0, so it must be at least 1, and since x_{1}+x_{2}+x_{3}+x_{4} = 1, we have that x_{2}=x_{3}=x_{4} = 0.

    So, we get N=1000 and \frac{N}{x_{1}+x_{2}+x_{3}+x_{4}} = 1000.

    Actually, N=2000,3000,4000,...,9000 works also, but in all those cases we have that \frac{N}{x_{1}+x_{2}+x_{3}+x_{4}} = 1000
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by pankaj View Post
    Let x_{1},x_{2},x_{3},x_{4} be the four digits of a four digit number N=x_{1}x_{2}x_{3}x_{4}.

    Find the maximum value of \frac{N}{x_{1}+x_{2}+x_{3}+x_{4}}
    N\ =\ 1000x_1+100x_2+10x_3+x_4

    \le\ 1000x_1+1000x_2+1000x_3+1000x_4

    \therefore\ \frac N{x_1+x_2+x_3+x_4}\ \le\ 1000

    Equality is attained when x_2=x_3=x_4=0. Hence the maximum value of N is 1000.
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  4. #4
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    Quote Originally Posted by TheAbstractionist View Post
    N\ =\ 1000x_1+100x_2+10x_3+x_4
    \le\ 1000x_1+1000x_2+1000x_3+1000x_4
    \therefore\ \frac N{x_1+x_2+x_3+x_4}\ \le\ 1000

    Equality is attained when x_2=x_3=x_4=0. Hence the maximum value of N is 1000.
    You mean the maximum value of \frac{N}{x_{1}+x_{2}+x_{3}+x_{4}} is 1000
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  5. #5
    Senior Member TheAbstractionist's Avatar
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    I do indeed.
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