In the multiplication problem above, A, B, and C represent distinct

A and B have to be a non-negative integer --> distinct digit
The fact that the question shows you the multiplication in a vertical display gives you a strong hint on how to solve it.
Neither A nor B can be zero --> B=0 then the product is 0, A=0 -->b=5 2A*B=100
Hence A,B are between 1-9
Using a vertical multiplication (which yields the same result as a horizontal multiplication, just a matter of convenience) the product of the factors' unit digits equals to the unit digit of their product.
If the product is lower then 10 then it only has a unit digit.
That is the case here as the sum of the two factors (in this case A+B) equals 5 and they are both<>0.
Therefore A*B=C
For the same reason the tens digit is equal to B*2
Hence 2*B=C
The third equation is given
A+B=5

Hope this helps

One thing I didn't simplify
A and B are between 1-9. since thier sum is 5 then they are between 1-4

The letters are placeholders for some specific digits. You need to plug in digits for the letters such that the multiplication relation holds. Also, a letter stands for the same digit in each occurrence i.e. the product is CC means both the digits of the product are the same.

In such questions, it is sometimes a good idea to look at the overall picture. A two digit number starting with 2 is multiplied by a single digit number to give CC i.e. 00 or 11 or 22 or 33 or 44 or 55 or 66 or 77 or 88 or 99
The only numbers out of these which can be obtained by multiplying 2A by B are 00 (If B = 0 but B and C must be distinct) or 22 (If A = 2 and B = 1 but A, B and C must be distinct), 44 (If A = 2 and B = 2 but again, A, B and C must be distinct) or 66 (If A = 2 and B = 3. Also, A + B = 5 so condition satisfied. This must be the answer.)

The multiplication will look like this:
__22
x__3
____
_66

Answer (A)

I solved this using algebra.
We are given:
AxB=C
& 2xB=C
& A+B=5
This yields to the following equation: (5-B)xB=2B --> 5B-B^2-2B=0 --> B(B-3)=0, hence B=0 or B=3
B can't be 0 because then C should be 0 and 0 isn't in the answer choices. B must be 3, A is then 2 and C is 6.

Hello Bunuel

Please I didn't get this question. if A= 2 and B= 3 the result will be 12 how did we concluded that c=6?

Thanks.

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If A = 2 and B = 3, we get 22 * 3 = 66, not 12

Bunuel. thanks for the reply.

Then, I think it is A*A not 2 A?

Best.

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2A there implies a two-digir number, where 2 is the tens digit and A is the units digit. The same for CC, it's a two-digir number, where both the tens digit and the units digit are C.­