ABC+BCB= CDD

In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?

a.8
b.10
c.12
d.14
e.18

ABC
BCB
CDD

To MAXIMIZE A and B, we must MAXIMIZE the value of C.

Case 1: C=9
AB9
B9B
9DD
Here, B=0, which violates the constraint that B is nonzero.

Case 2: C=8
AB8
B8B
8DD
Here, B=1, implying that A=7.
In this case, A*B = 7*1, which is not among the answer choices.

Case 3: C=7
AB7
B7B
7DD
Here, it's possible that B=2, implying that A=5.
In this case, A*B = 5*2 = 10.

Case 4: C=6
AB6
B6B
6DD
Here:
If B=3, then A=3, in which case A*B = 3*3 = 9.
If B=2, then A=4, in which case A*B = 4*2 = 8.
If B=1, then A=5, in which case A*B = 5*1 = 6.

At this point, we can see that the maximum possible product is yielded by Case 3:
A*B = 5*2 = 10.

The correct answer is B.

prernamalhotra wrote:ABC
+BCB
___
= CDD

In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?

1) 8
2) 10
3) 12
4) 14
5) 18


Thank you,
Prerna

Gmatguru's analysis might be the first approach that comes to mind. But here's another way to look at it.

As per the statements provided, Since A,B,C and D are non zero single digits, we can arrive at these equations.

A+B=C
B+C=D

Combining these 2 we get, A+2B=D
D's Maximum possible would be 9.

So, A+2B=9
Possible values here for A and B are (7,1) (5,2) (3,3) (1,4)
10 is the only option available and thats the maximum.

Hope this helps!!