In the addition shown above, A, B, C, and D represent the nonzero digi

I'm happy to help with this.
First, look at the one's column. C + B produces a one's unit of D, and we don't know whether anything carries to the ten's place.
But now, look at the ten's place --- B + C again produces a digit of D --- that tells us definitively that nothing carried, and that B + C = D. B & C have a single digit number as a sum.
Now, C = A + B, because again, in the hundreds column, nothing carries here.

Notice that, since C = A + B and D = B + C, B must be smaller than both B and C. We want to make the product of A & B large, so we want to make those individual digits large. Well, if we make B large, then that makes C large, and then B + C would quickly become more than a one-digit sum, which is not allowed. Think about it this way. Let's just assume D = 9, the maximum value.
D = 9 = B + C = B + (A + B) = A + 2B
We want to pick A & B such that A + 2B = 9 and A*B is a maximum. It makes sense that B would be smaller.
Try A = 7, B = 1. Then A + 2B = 9 and A*B = 7
Try A = 5, B = 2. Then A + 2B = 9 and A*B = 10
Try A = 3, B = 3. Then A + 2B = 9 and A*B = 9
Try A = 1, B = 4. Then A + 2B = 9 and A*B = 4
Indeed, as B gets bigger, the product gets less. This seems to imply that the biggest possible product is 10. This corresponds to A = 5, B = 2, C = 7, and D = 9, and the original addition problem becomes
527
+272
799
Thus, the maximum product is 10, and answer = (B).

Does all this make sense?
Mike
_________________

This problem can be solved much faster if you start with the options. As Mike has wonderfully explained, none of the digits will have a carry over.

Option E is \(18 = 9 * 2\)(no other case is possible). Now\(9+2>=10\). Therefore, it will have a carry over digit 1. So reject this option.

Option D is \(14 = 7 * 2\)(no other case is possible). Since \(C=A+B\), therefore \(C= 9\). Since \(C=9\) and \(B\) is a non zero digit, \(C+B>=10\). Therefore, it will have a carry over digit 1. So reject this option.

Option C is 12. This will have 2 cases-
\(Case 1--> 12 = 4 * 3\)

Here \(C= 4+3=7\). Now \(7+4>=10\) and [\(7+3>=10\). Therefore, both cases will have a carry over digit 1. So reject this option.


(OR) \(Case 2--> 6 * 2\).

Here \(C= 6+2=8\). Now \(8+2>=10\) and \(8+6>=10\). Therefore, both cases will have a carry over digit 1. So reject this option.

Option B is \(10 = 5 * 2\). If you follow the same set of steps you'll notice that there will not be any carry over digit for the case \(C=7, A=5, B=2\). So right answer!

Hope it was helpful.

How do we do all that in less than 2 mins, is my only question

Posted from my mobile device

Dear Saint For Life,
Problems such as this are quite out-of-the-box. To some extent, the GMAT intends us to handle many of the other questions in 90 secs or less, so that we have a bit of a time-cushion when we run into one of these oddball questions.

Having said that, the more familiar you are with number properties, the faster it will go. In this problem, it took me quite some time to write out everything in verbal form, but I saw things relatively quickly. As you practice seeing patterns, you will see them more quickly, even if they are hard to explain to someone else. You may find this post helpful:
https://magoosh.com/gmat/2013/how-to-do- ... th-faster/

Mike
_________________

Similar questions to practice:
in-the-correctly-worked-addition-problem-above-a-b-c-d-128953.html
tough-tricky-set-of-problems-85211.html#p638336
in-the-correctly-worked-addition-problem-shown-where-the-54154.html

Hope it helps
_________________

... got E as an answer. However, it took me 6 minutes, since I plugged in the answer options.
Is there any faster way, or am I just too slow in back-solving?



My explanation:

According to the question stem C+B=D and D<10, since the second addition (2nd column) is C+B=D again (if there were a carry, the second column's result would not be D again), and A+B=C
Now I break up the answer options:
A) A*B=8, so A and B -> 2 and 4 or 1 and 8. A+B=2+4=6=C, C+B=6+2=8=D, which is smaller than 10, thus the answer is not "out", but there could be a larger possible value of A*B
B) A*B=10, so A and B -> 2 and 5. A+B=2+5=7=C, C+B=7+2=9=D, which is smaller than 10, thus the is also possible. Let's look for the next options.
For C-E:
C) A*B=12, so A and B -> 6 and or 4 and 3.
D) A*B=14, so A and -> 7 and 2.
E) A*B=18, so A and B -> 9 and 2 or 3 and 6.
Do the same process with those numbers and you will find that all will yield a sum of C+B>10, thus the constraint of D<10 is not satisfied. The answer options are not possible.
Because 10>8 answer option B) is the largest possible.

I do a thorough search before posting any new question. But this time I couldn't find this link
Thanks for sharing the link
Forum moderator may delete this thread

Wonderful explanation. Just thought to add some more from my side. As we see only B satisfies all the necessary requirement(not to carry over any digits).

In option A, taking A=4 and B=2 also satisfies the condition as it is something like
426
+262
-------
688

But since we need maximum value well go for option B

Dear Mike, I was wondering why are we maximising D since it no where mentioned to do so. We have to maximise A*B right so if we choose C=1 and B=5 and A=4 i think we staisfy all conditions and our product of A*B (5*4) comes to be 20. Dont know where i am going wrong. Kindly help

Dear Sarthak.bhatt,

I'm happy to respond.

With all due respect, my friend, you are not interpreting the question correctly. There is absolutely no multiplication happening in this question. Here's the prompt again:

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 ?

Thus, for example if C = 1, B = 5, and A = 4, then ABC would not be the product of those three numbers but instead the single three-digit number 451. With those numbers, the problem would be

451
+515
966

These choices satisfy the equation.

The problem is completely different from the way you were conceptualizing it, so the strategy is completely different from what it would be in the problem you had in mind.

Does this make sense?

Mike
_________________

mikemcgarry. Can you please explain Sarthak's query as i have the same query but unfortunately i didn't get from the reply you gave to him.

Do we not consider negative numbers in this problem at all?

No. We are given that A, B, C, and D represent the nonzero digits. There are 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
_________________

it took me "just" 5 minutes... I understand the cushion but it's a killer question for most of us.

Guys can you please elaborate why we did'nt chose option E
A=6 and B=3, which make product (highest) 18

ABC 631
+BCB +313
CDD =944

Why it is wrong??

I approached to this question as below:

Given in the QN

1. B+C = D ; Where D <= 9 ( Derived from unit and tenth digit )
2. A+B = C ; Derived from 100th digit

From 1 & 2, we can say B+(A+B)=9 (taking maximum value)
Thus: A+2B = 9 and A=9-2B (Equation 3) and 2B < 9 i.e. B <4 ( as it's integer)


From eqn 3, we can put A & B to find the required values.

B=1 , A = 7, Thus AB = 7
B=2 , A = 5, Thus AB = 10
B=3 , A = 3, Thus AB =9
B=4 , A = 1, Thus AB = 4

So, Answer is 10

Hello SameerGupta2001,

In the above example that you have provided, you have established C=1 but afterwards C becomes 9 , which is wrong , because C must have a single value, but in the above example 6+3=9 , while in the beginning you have put C=1.

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


_________________

In this sort of a question, it seems obvious that A, B, C, D are all different digits, but is it alright to assume that? The question doesn't mention that they are different numbers. What if the total is 999, with A = C = 5 and B = 4. Then ABC + BCB becomes 454 + 545, adding up to 999 (where C and D are both 9) and AB = 20.