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Thread: Is B bigger than A?

  1. #1
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    Is B bigger than A?

    I know that

    | 191*A = 70*B + 898*C
    | 70*B = 191*A - 898*C
    | 898*C = 191*A - 70*B

    I can say for sure that C<A
    But can I say B>A without guessing ?
    (A,B,C are natural numbers 0-9)
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  2. #2
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    Lightbulb Re: Is B bigger than A?

    Quote Originally Posted by Phairero View Post
    I know that

    | 191*A = 70*B + 898*C
    | 70*B = 191*A - 898*C
    | 898*C = 191*A - 70*B

    I can say for sure that C<A
    But can I say B>A without guessing ?
    (A,B,C are natural numbers 0-9)
    ... But natural no.s start from 1
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  3. #3
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    Re: Is B bigger than A?

    Well, some include 0 too.
    What I meant was they are ... numbers from 0-9
    as in (0,1,2,3,4,5,6,7,8,9)
    (Edit (added) : NOT 1,3 or pi or 3,89 or something else like that)
    What were they called... oh, one-digit numbers!
    Sorry my bad.
    Can you help me now?
    Last edited by Phairero; Oct 23rd 2011 at 02:01 AM. Reason: Corrected a typo. Added some info.
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  4. #4
    Member sbhatnagar's Avatar
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    Lightbulb Re: Is B bigger than A?

    Our equation is $\displaystyle 191A=70B+898C$

    Part 1)

    The equation can be rearranged to form $\displaystyle -898C=70B-191A$

    We know that $\displaystyle 0\leqslant C \leqslant 9$ (C is positive, so a positive no times a negative no is negative), so LHS is negative & $\displaystyle B $ and $\displaystyle A$ are positive so this leads us to the fact that $\displaystyle 191A>70B$.

    Part 2)

    The equation can be rearranged to form $\displaystyle -70B=898C-191A$

    Again here, $\displaystyle 191A>898C \Leftrightarrow A>C$
    Last edited by sbhatnagar; Oct 23rd 2011 at 03:42 AM.
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  5. #5
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    Re: Is B bigger than A?

    Yes, I apologize If I can't read it out, but where does it say then, that B > A ?
    Can I say it?

    I already know the answer I just need to know if I can say that B is bigger than A
    If I can't say it my whole solution might just fall apart.
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  6. #6
    Member sbhatnagar's Avatar
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    Lightbulb Re: Is B bigger than A?

    You may or may not say that $\displaystyle B>A$ because we found in part 1 that $\displaystyle 191A>70B$. This does not tell us whether $\displaystyle A>B$ or $\displaystyle B>A$.

    In part 2 we found that $\displaystyle 191A>898C$ but here $\displaystyle 898>191$ so we can easily say that $\displaystyle A>C$.
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  7. #7
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    Re: Is B bigger than A?

    That is sad...
    Then how am I supposed to figure this out

    1998*ABC=CBA*8991

    ABC = 3 digit number
    CBA = same 3 digit number written backwards
    A,B,C are one digit numbers.

    What I did until that.

    1998 * ABC = CBA * 8991
    1998 * (100A + 10B + 1C) = (100C + 10B + 1A) * 8991
    199800A + 19980B + 1998C = 899100C + 89910B + 8991A
    199800A - 8991A = 89910B - 19980B + 899100C - 1998C
    190809A = 69930B + 897102C |divided by 9
    21201A = 7770B + 99678C |divided by 3
    7067A = 2590B + 33226C |divided by 37
    191A = 70B + 898C

    191 is a prime number, so I can't divide any further.
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  8. #8
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    Re: Is B bigger than A?

    Hello, Phairero!

    Solve the alphametic: .$\displaystyle 1998\cdot ABC\,=\,8991\cdot C\!B\!A$

    $\displaystyle ABC$ = 3-digit number
    $\displaystyle C\!B\!A$ = same 3 digit number written backwards
    $\displaystyle A,B,C$ are digits.

    We have: . . .$\displaystyle 1998\cdot ABC \:=\: 8991\cdot C\!B\!A$

    Divide by 999: w$\displaystyle 2\cdot ABC \:=\:9\cdot C\!B\!A$

    . . . .$\displaystyle 2(100A + 10B + 1C) = 9(100C + 10B + 1A)$

    . . . . . . $\displaystyle 200A + 20B + 2C = 900C + 90B + 9A$

    . n . . . . . . . . . . . . $\displaystyle 191A \:=\:70B + 898C$ .[1]

    . n . . . . . . . . . . . . $\displaystyle 191A \:\equiv\: 898C\text{ (mod 70)}$

    . . . . . . . . . . . . . . . $\displaystyle 51A \:\equiv\:58C\text{ (mod 70)}$

    Multiply by 11:. . . . $\displaystyle 561A \:\equiv\:638C\text{ (mod 70)}$

    . w . . . . . . . . . . . . . . $\displaystyle A \:\equiv\: 8C\text{ (mod 70)}$


    Since $\displaystyle A$ and $\displaystyle C$ are digits: .$\displaystyle A \,=\,8,\;C\,=\,1$

    Substitute into [1]: .$\displaystyle 191(8) = 70B + 898(1)$

    . . Hence: .$\displaystyle B \,=\,9$

    Therefore: .$\displaystyle \boxed{ABC \:=\:891}$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Check:. . $\displaystyle \begin{Bmatrix}1998 \times 891 \\ 8991 \times 198 \end{Bmatrix} \;=\;1,\!780,\!218$

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  9. #9
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    Re: Is B bigger than A?

    I did some research, and understood most of it, except how you got from

    to


    EDIT :

    And from this


    to


    or can I just write them there and every teacher/tutor/professor/smarter-person-than-me would understand me?
    Last edited by Phairero; Oct 23rd 2011 at 06:33 AM. Reason: one more thing
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