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Math Help - An Easy-Hard Question

  1. #1
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    An Easy-Hard Question

    The mystery # is a decimal....

    -The digit in the 1's place is less then then digit in 10th place..
    -No 2 digits are the same..
    -The number is less than the # of years in a century...
    -The digit in the 10th place is less than the digit in the 1000's...
    -There are 2 digits in the whole part..
    -There are 5 digits all together...
    -The sum of the digits is 20....
    -No digit is odd...
    -The product of the digits in the ones place and the 100th is zero...
    -The sum of the digits in the ones place and the 100th is two...
    -The number is greater than 5 dozen...
    -The sum of the digits in the decimal part of the # is the same as the digits in the whole part of the number.
    -The product of the digits in the 10ths place and the 1000's is equal to 2 dozen...
    -The product if the digits in the 100's place and the tens place is zero..

    -What is the mystery # ________

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  2. #2
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    Let us begin with these pieces:
    - The number is less than the # of years in a century...
    - There are 2 digits in the whole part..
    - There are 5 digits all together...
    - The number is greater than 5 dozen...

    From this given, we can establish that:

    x = 10a + b + 0.1c + 0.01d + 0.001e

    60<x<100

    Now, let us translate the rest of the given into algebraic terms:

    b<c

    c<e

    a+b+c+d+e=20

    bd=0

    b+d=2

    c+d+e=a+b

    ce = 24

    cd=0

    1. Now, step number 1: solve simultaneously:

    a+b+c+d+e=20

    a+b-c-d-e = 0

    to get a+b = 10

    2. Now solve simultaneously again with:

    a+b = 10

    b+d = 2

    to get a-d = 8

    3. Then, let us examine the given:

    ce = 24

    cd=0

    It is obvious to see that d = 0. It is impossible that c = 0 since the product of c and e is 24. This is the key; plug in the relations we got previously:

    4.
    \mbox{d = 0}

    \mbox{a = 8}

    \mbox{b = 2}

    a+b+c+d+e=20

    \implies c+e = 10

    \implies c+\frac{24}{c} = 10

    \implies c^2 - 10c + 24 = 0 \implies (c-4)(c-6)=0

    If c = 6, then e = 4 but this doesn't satisfy the inequality c < e
    Therefore, c = 4 and e = 6

    \therefore x = 10(8)+2+0.1(4)+0+0.001(6) = 82.406
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  3. #3
    MHF Contributor arbolis's Avatar
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    I'm getting all confused with "whole part" and "all together" and also with digit.
    So I didn't bother to solve the problem. Anyway, I have a question from
    -The digit in the 10th place is less than the digit in the 1000's...
    and
    82.406
    Are there 10 places for digits? 1000? In 82.406 of course.
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  4. #4
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    The whole part usually refers to the part of the number that is not a decimal/ left to the decimal point. Take 123.45 for instance:

    123.45 = 1(100) + 2(10) + 3(1) + 4(0.1) + 5(0.01)

    123 is the whole part. 1 occupies the hundred place, 2 occupies the ten place, 3 occupies the one place, 4 occupies the tenth place, and so on. To refer to a place that is left of the decimal, we follow the same naming process as we did to digits to the left of the decimal but we add the suffix -th to indicate a fraction.

    \text{Tenth} \implies \frac{1}{10}

    \text{Hundredth} \implies \frac{1}{100}

    \text{4~Hundredths} \implies \frac{4}{100}

    For 82.406, there is a tens place. There is no 1000 place, but there is a 1000th place.
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  5. #5
    MHF Contributor arbolis's Avatar
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    Thank you Chop Suey.
    I'm still confuse about
    -The digit in the 10th place is less than the digit in the 1000's...
    should 1000's be replaced by 1000th? Hmm, maybe not. I don't understand well, nevermind.
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  6. #6
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    Quote Originally Posted by arbolis View Post
    Thank you Chop Suey.
    I'm still confuse about should 1000's be replaced by 1000th? Hmm, maybe not. I don't understand well, nevermind.
    Actually, you are right. The 1000's should be replaced by 1000th. The problem may not have been copied in a consistent manner.
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