&, #, and each represents a different digit, and & multiplied by #

Let the digits be a,b and c. We have to find \(ab\).
It is given that
\(ab\) x \(ba\) = \(bcb\)

In the product,
b at unit's place will be obtained by a*b
c at ten's place will be obtained by a*a+b*b
b at hundred's place will be obtained by b*a

Since b*a is b, a must be 1.
So,
\(1b\) x \(b1\) = \(bcb\)

c is a*a+b*b or 1*1 + b*b ......(1)
Notice that b at hundred's place is obtained by b*1 and has no carry over from (1).
Thus b can take only be equal to 1 or 2 as there must be no carry over.
The question says that all digits are different , so b must be 2 as a is 1.
so our digit is 12
Answer:- B

Could you explain why Option D is wrong ?

Guess both D and B fit in as valid answers.. The options have not been given correctly.

We are asked to find value of &#.
Look at the product &#*#& = #@#
& on multiplication with # gives #. This is only possible if & is 1 as it is given that product of & and # is smaller than 10.
If option D is chosen, then the product would look like this &#*#& = &@&
Hence, & cannot be 2 and our number cannot be 21.

VERITAS PREP OFFICIAL SOLUTION:


At first glance, this problem looks like nothing you’ve seen before! But you do know some things here. When multiplying two-digit numbers, the first step is to multiply the units digits. And here, # * & provides a units digit of #. How is that true, particularly if as stated above the full product is less than ten? That means that & must equal 1, so that when it multiplies with another number, that other number (#) stays the same. 1 * # = #, so we can prove that & is 1.

We also know that the digits must be unique, so choice A is eliminated, and the fact that the & is 1 means that D and E are eliminated (the first digit must be a 1). So that leaves just two options, 12 and 13. And we know that the problem, then, is either 12*21 or 13*31, and we have to have a solution with the units and hundreds digits the same. At this point, we have two quick multiplication problems to do:

12*21 = 252 (this works)

13*31 = 403 (this does not)

So the correct answer must be 12. The key is to begin with a first step, taking inventory and advantage of what you know and working from there. Difficult problem solving questions often hinge on the exam’s knowledge that people are uncomfortable when they cannot see the entire process all at once, but remember that they also have to be written such that many examinees can solve them in 2 minutes. So rest assured that even if you do not know much, you will still know enough to get started. Great test takers always begin by taking what they know and working from there; those who don’t score as well tend to focus more on what they don’t know, and struggle to get started.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.