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Math Help - need help please!!!

  1. #1
    dlhenry
    Guest

    need help please!!!

    Ann A. Bel, a bright mathematices student, has discovered something interesting about her name. If the letters are arranged as they are below, it is possible to replace each different letter with a different digit and have the multiplication work out correctly. What digit should replace each letter? Be sure to expain your work

    ANN
    x A
    BEL

    If you could help me out I would really appreciate this.

    Thanks
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
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    Hello, dlhenry!

    This takes simple reasoning ... but a lot of it!


    Solve the alphametic:
    Code:
          A N N
          x   A
          -----
          B E L
    Consider the possible values for A.

    A \neq 0; the product would be zero.

    A \neq 1; the product would be ANN.

    A \neq 4,5,6,7,8,9; the product would be a four-digit number.

    . . Hence, A \,= \,2,\:3


    Suppose A = 3.
    Then we have:
    Code:
          3 N N
          x   3
          - - -
          9 E L

    In the leftmost multiplication, we have: . 3 \times 3 \,=\,9
    . . There is no "carry" from the second multiplication, 3 \times N
    Hence: . N \:=\:0,\,1,\,2

    But none of these will work.
    . . If N = 0, we have: . 300 \times 3 \:=\:9{\color{red}00}
    . . If N = 1, we have: . 311 \times 3 \:=\:9{\color{red}33}
    . . If N = 2, we have: . 322 \times 3 \:=\:9{\color{red}66}
    In all cases, E\,=\,L . . . which is not allowed.


    Therefore: . {\color{blue}\boxed{A = 2}}
    And we have:
    Code:
          2 N N
          x   2
          - - -
          B E L

    In the rightmost mutliplication: . 2 \times N ends in L.

    In the second multiplication: . 2 \times N ends in E, a different digit.

    Hence, there must be a "carry" from the first multiplcation.
    . . So, N \:=\:5,\,6,\,7,\,8,\,9


    Since the second multiplication will also have a "carry",
    . . the leftmost multiplication is: . (2\times2) + 1\:=\:5

    Therefore: . {\color{blue}\boxed{B \,=\,5}}


    So: . N \:=\:6,\,7,\,8,\,9

    If N = 6, we have: . 266 \times 2 \:=\:53{\color{red}2}
    . . which makes L = 2 . . . and 2 is already used.

    If N = 7, we have: . 277 \times 2 \:=\:5{\color{red}5}4
    . . which makes E = 5 . . . and 5 is already used.

    If N = 9, we have: . 299 \times 2 \:=\:5{\color{red}9}8
    . . which makes E = 9 . . . and 9 is already used.


    The only remaining choice is: . {\color{blue}\boxed{N \,= \,8}}

    Therefore, the solution is:
    Code:
          2 8 8
          x   2
          - - -
          5 7 6
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