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Question

Find the digit A
   B A×B 3   57A


  1. 2

  2. 5

  3. 3

  4. 4


Solution

The correct option is B

5


Ones digit of 3 × A is A. So, it must be either A = 0 or A = 5.

Now, if B = 1, then BA × B3 would at most be equal to 15 × 13; (Taking A= 5 to find the maximum value of the product) that is, it would at most be equal to 195. But the product here is 57A, which is more than 500. So we cannot have B =1.

If B = 3, then BA × B3 would be at least 30 × 33; (Taking A = 0 to find the minimum value of the product) that is 990. But 57A is less than 990. So, B cannot be equal to 3.

Putting these two facts together, we get B = 2.
So, the multiplication is either 20 × 23, or
25 × 23.

The first possibility fails, since
20 × 23 = 460. But, the second one works out correctly, since
25 × 23 = 575. So the answer is A = 5 and B = 2. 
   2 5×2 3   575

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