A Cryptarythmic Tribute To Bob Marley

Let us remind ourselves of the code:

   BOB
x  BOB
------
  MEOY
 MILO
MEOY
------
MARLEY

Let us notice the second 'solution line' [I dunno what's it called, really.]. In the line $MILO$ which would represent the product of the first $BOB$ by the $O$ of the second $BOB$ , we know that the last digit is represented by an $O$ . That $O$ represents the units digit of the product of the second $B$ of the first $BOB$ and the $O$ of the second $BOB$ , that is, the last digit of the product of $B$ and $O$ has a last digit of $O$ . There are a lot of possibilities here as I will show in the table below:

       B x O = _O
-----------------
       n x 0 =  0
       1 x n =  n
       6 x 2 = 12
       6 x 4 = 24
(2m + 1) x 5 = _5
       6 x 8 = 48

The value $n$ represents any number [recall your identity and zero properties of multiplication]. The value $2m + 1$ is an odd number: $1 \times 5$ will give you $5$ , $3 \times 5$ will give you $15$ , and so on.
We will use these information to look at other parts of the crypt and deduce the values of the digits.

If we look at the product of the last $B$ s of $BOB$ , we know that $B \times B$ has a last digit of $Y$ , which means that the last digit of the product of $B \times B$ is not also $B$ . From here, we can rule out the possibility of $B$ being a $0$ , a $1$ , or a $6$ since $0 \times 0 = \textbf{0}$ , $1 \times 1 = \textbf{1}$ and $6 \times 6 = 3\textbf{6}$ and the remaining options are

       B x O = _O
-----------------
       n x 0 =  0 [except n = 0, 1, 6]
(2m + 1) x 5 = _5 [except 2m + 1 = 1]

Looking back at the product $B \times B = Y$ , knowing that $B$ is not $0$ , $1$ or $6$ , we have the possible products:

$\begin{aligned} 4 = 2 \times 2 \\ 9 = 3 \times 3 \\ 16 = 4 \times 4 \\ 25 = 5 \times 5 \\ 49 = 7 \times 7 \\ 64 = 8 \times 8 \\ 81 = 9 \times 9 \end{aligned}$ .

With the exception of the possibilities $B = 2$ and $B = 3$ , the product $B \times B$ will yield a two-digit number. If that's the case, then the tens digit of the product of the last $B$ s of $BOB$ will be added to the product of $O \times B$ . But from what we knew about the product of the second $B$ of the first $BOB$ and the $O$ of the second $BOB$ which has a last digit of $O$ , we should deduce that the product of $B \times B = Y$ should be a one-digit number: had it been a two-digit number, then whatever the tens digit of $B \times B$ will be, that will be added to the product of $O$ of the first $BOB$ and the last $B$ of the second $BOB$ and the result will bear some other last digit and not $O$ . Thus, the possibilities of $B$ being either $4$ , $5$ , $7$ , $8$ , or $9$ should be discarded, leaving us with two possibilities for $B$ : $B = 2$ and $B = 3$ . Our candidates table has now shrunk to

       B x O = _O
-----------------
       2 x 0 =  0
       3 x 0 =  0
       3 x 5 = 15

Finally, observe the second adding column from the right:

   BOB
x  BOB
------
  MEOY
 MILO
MEOY
------
MARLEY
    ^

Since $B \times B$ has a single-digit product, $Y$ , nothing is to be carried over $O + O = E$ . Now, $O + O = E$ implies that the last digit of the sum of two $O$ s is not also an $O$ . We can discard the possibility of $O = 0$ , leaving us with

       B x O = _O
-----------------
       3 x 5 = 15

That is, $B$ is the digit $3$ and $O$ is the digit $5$ . Thus, $BOB$ is actually $353$ . The rest is just calculating the product of $353 \times 353$ :

   353
x  353
------
  1059
 1765
1059
------
124609

The sought answer is $\boxed{353124609}$ .

1.B has to be either 2 or 3 because in both 1st n 2nd step of multiplication, O*B=O (ie B^2 has to be a single digit).

2.O*B results in unit digit O only if O is equal to 5 and B is equal to 3. (O can never be 0)

Rest of the sum can be easily solved.