Write this problem in a vertical multiplication format:

SAY

3

------

BABY

Start multiplying from right to left, as usual for vertical multiplication. When you multiply Y (a one-digit number) by 3, you must get either Y or a two-digits number that ends with Y. This is only possible when Y = 0 or Y = 5.

Consider first the case when Y = 0. The problem then becomes

SA0

3

------

BAB0

Now, notice that B can only be 1 or 2, because it is impossible to produce a number larger than 30 when multiplying a single-digit number (S) by 3. Even if there was 1 or 2 carried over from multiplication of A by 3, it would be impossible.

Let's see what happens if B = 1. If B = 1, A times 3 must be a number that ends with 1. This is only possible when A = 7. Then, the problem becomes

S70

3

-----

1710

Is this possible? Yes, because 7*3 = 21, which means there is 2 carried over, so multiplication of S times 3 has to be 17 - 2 = 15. This is possible when S = 5. So, we have an answer

**S = 5, A = 7, Y = 0, B = 1.**

Let's see if there are other possible answers. If Y = 0 and B = 2, then

SA0

3

------

2A20

would have to work. This can happen when A = 4, because 4*3 = 12, and 1 is carried over. But then the problem becomes

S40

3

-----

2420

and S times 3 must be 24 - 1 = 23, which is impossible, because 23 is not divisible by 3. So Y = 0 and B = 2 does not work.

We can also consider cases when Y = 5, B = 1 and Y = 5, B = 2. By trying out the possible values of A, as done above, it can be shown that there are no possible values of S. So Y = 5 will not work and there is only one solution:

**S = 5, A = 7, B = 1, Y = 0.**