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Thread: 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 $\displaystyle A.$

    $\displaystyle A \neq 0$; the product would be zero.

    $\displaystyle A \neq 1$; the product would be $\displaystyle ANN.$

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

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


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

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

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


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

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

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

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


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

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


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

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

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

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


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

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