Hello, dlhenry!

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

Consider the possible values forSolve the alphametic:

Code:A N N x A ----- B E L

; the product would be zero.

; the product would be

; the product would be a four-digit number.

. . Hence,

Suppose .

Then we have:Code:3 N N x 3 - - - 9 E L

In the leftmost multiplication, we have: .

. . There is no "carry" from the second multiplication,

Hence: .

But none of these will work.

. . If , we have: .

. . If , we have: .

. . If , we have: .

In all cases, . . . which is not allowed.

Therefore: .

And we have:Code:2 N N x 2 - - - B E L

In the rightmost mutliplication: . ends in

In the second multiplication: . ends in , a different digit.

Hence, there must be a "carry" from the first multiplcation.

. . So,

Since the second multiplication will also have a "carry",

. . the leftmost multiplication is: .

Therefore: .

So: .

If , we have: .

. . which makes . . . and 2 is already used.

If , we have: .

. . which makes . . . and 5 is already used.

If , we have: .

. . which makes . . . and 9 is already used.

The only remaining choice is: .

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