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Math Help - find the number

  1. #1
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    Exclamation find the number

    Heres the problem:
    The tens digit of a two-digit number is 1 more than 3 times the ones digit. Subtracting 45 from the number reverses it. Find the original number.
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by sp0rtskid55 View Post
    Heres the problem:
    The tens digit of a two-digit number is 1 more than 3 times the ones digit. Subtracting 45 from the number reverses it. Find the original number.
    well,
    If the 1's digit is 0, then 3*0+1 = 1, so the number is 10
    If the 1's digit is 1, then 3*1+1 = 4, so the number is 41
    If the 1's digit is 2, then 3*2+1 = 7, so the number is 72
    If the 1's digit is 3, then 3*3+1 = 10, this is not a two digit number, so you are left with the first three cases.

    Just check which one meets the criteria of "Subtracting 45 from the number reverses it"
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  3. #3
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    Hello, sp0rtskid55!

    Let t = tens digit
    Let u = ones digit

    The number is: . 10t + u

    Its reversal is: . 10u + t


    The tens digit of a two-digit number is 1 more than 3 times the ones digit.
    {\color{blue}t \:=\:3u+1}

    Subtracting 45 from the number reverses it.
    {\color{blue} 10t + u - 45 \:=\:10u + t}

    Find the original number.

    The second equation is: . 9t\:=\:9u + 45\quad\Rightarrow\quad t \:=\:u + 5

    Equate to the first equation: . 3t + 1 \:=\:u + 5\quad\Rightarrow\quad u \:=\:2
    . . Then: . t = 7


    Therefore, the original number is 72.

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