
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

Hello, dlhenry!
This takes simple reasoning ... but a lot of it!
Quote:
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 fourdigit 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