• Sep 9th 2007, 01:17 PM
dlhenry
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
• Sep 9th 2007, 02:43 PM
Soroban
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 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