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

Question

&, #, and @ each represents a different digit, and & multiplied by # is smaller than 10. What is the value of the two-digit number &# if the following multiplication problem is true:

&#*#& = #@#

(A) 11
(B) 12
(C) 13
(D) 21
(E) 31

sunnysid · Answer

The simplest approach for me was to test the answers choices.

Given: &# * #& = #@#

Testing the answer choices as per the above condition:

A) 11*11 = 121
B) 12*21 = 252
C) 13*31 = 403
D) 21*12 = 252
E) 31*13 = 403

Since the nos. &,#, @ are different we can eliminate 11.
Out of the others only option B satisfies the condition-> &# * #& = #@#

Ans- B

GMATinsight · Answer

Given: & multiplied by # is smaller than 10

i.e. & and # are 2 and 3 (Not essentially in that order)
OR
i.e. & and # are 1 and 2 (Not essentially in that order)
OR
i.e. & and # are 1 and 3 (Not essentially in that order)

None of the options have (2, 3) both digits so it's not needed to be checked

Checking &#*#& = #@#   for (1, 2)

12*21 = 252 (i.e. of the form #@#) Hence correct answer

Answer: option B

kunal555 · Answer

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

rohanadagouda · Answer

Could you explain why Option D is wrong ?

itwarriorkarve · Answer

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

kunal555 · Answer

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.

Bunuel · Answer

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.

bumpbot · Answer

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

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