Other than 10

By elimination of impossible choices we will arrive at the result. 1) The digit in the unit place has to be 1 or 0 only. As any other digit will leave the same digit in product's unit place. 2) If 0 is the digit in unit place, any digit other than 1 in the 10th place will give the same digit in the last place. But that is not acceptable, so 0 cannot be in unit place. 3) 1 has to be in the unit place. By enumerating between 11 and 91 for multiplying with just 11, we find that all but 91 are not valid.

Let the two digit number be ab. If the product of the number ab with 11 only contains 1s and 0s then it will hold good for other numbers too.

So (10a+b)11 =110 * a+11 * b=aa0+bb. (Note-a and b are single digit numbers) 110a definitely ends with 0. So ‘b’ can not be any single digit number except 0 or 1.(As 11 * b=bb) Let’s say b is 0. Possible values are 20,30,40….90. None of them produce desired result. Hence, b=1. So (aa0+11) must contain only 1s and 0s where ‘a’ can be (1,2,3…9) None of the value satisfies the above expression to get the result except 9. (990+11=1001). So a=9 Hence, the desired two digit number is 91.