Find the sum of the reciprocals 
Let $A$ be the sum of the reciprocals of the positive integers that can be formed by only using the digits $0,1,2,3$. That is, $$A = \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+\dfrac{1}{30}+\cdots.$$ What is $\lfloor A\rfloor$?

The sum is greater than $2$ since $\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11} > 2$. Can we show it is less than $3$?

Although this is clearly a faulty approach, take the first 15 formable numbers. 
These are 1,2,3,10,11,12,13,20,21,22,23,30,31,32,33. 
The decimal reciprocals are roughly equal to $1+0.5+0.35+0.1+0.09+0.08+0.08+0.05+0.05+0.04+0.04+0.03+0.03+0.03+0.03$, more or less by a very small amount.
This is slightly less than 2.5
Using three digits, you can form somewhere around (4)(4)(3)=48 numbers.
Because these are 3 digit numbers, the reciprocal of each must be less than 0.01. About 0.004 on average.
So adding this, the sum is way less than 2.98 (closer to 2.7) because only 1/3 of the denominators are closer to 100.
Using 4 digits, we can create (4)(4)(4)(3)= 192 numbers. Each reciprocal is less than 0.001. About 0.0004 on average.
The final sum is about 2.841.
