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Not sure, but this is how I solve it:
X=ab ( a=tens, b=ones)
(10*a )+ b=3 (a+b)
a=7/2 b, if x is two digit, b should be 2 to make a less than 10, therefore b=2 and a=7
x= 72

Not sure, but this is how I solve it: X=ab ( a=tens, b=ones) (10*a )+ b=3 (a+b) a=7/2 b, if x is two digit, b should be 2 to make a less than 10, therefore b=2 and a=7 x= 72

Correction:
a = 2/7b

I think now you'll see that Statement (2) doesn't do anything for us.

My mistake, a=2/7 b but shouldnтАЩt тАЬbтАЭ be 7 for a to be less than 10 Can you elaborate more about your answer? Thanks

I think Statement (1) is sufficient because 27 is the only such integer. I tried to find solution for 3 and 4 digit numbers before posting and wasn't able to come up with anything. In fact, I would hit dead end after two iterations.

IMO, this is a really bad question. Whoever generated it. One must know something so whacky as this to solve the problem. But I give props to sarnia for taking the right approach in solving it.

Case 1: X cannot be a in single digits. Obviously.
Case 2: Say x has 2 digits, then 10a+b=3(a+b)=>a/b=2/7. Since a and b have to be single digits and not equal to 0, the only option is a=2,b=7.
Case 3: Say x has 3 digits, the maximum value can be 999, sum of digits = 27. Multiply with 3 to get 81. Even the maximum of the sum will not get near the minimum of 3 digit number 100. Therefore X cannot be a 3 digit number.

The logic in Case 3 can be extended for numbers with more than 2 digits (4,5 etc.). Therefore the answer is A.

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