Amount of amplification required for a certain accuracy We want to measure the resistance of a resistor up to a certain accuracy. To this end, we use a fixed voltage and measure the current through a I/V converter. This measured voltage is then sent to an A/D converter with a certain input voltage range.
Now, what I'm puzzled about that the accuracy of our measurement can be affected by the amplification of our I/V converter. I would suspect that once we've set the I/V converter to the right amplification value, the accuracy would be completely determined by the resolution of our A/D converter.
For example, say we want to measure a resistance R using a voltage V. Our current would then be I = V/R. If we now want our measurement to be up to, say, 1 percent, what kind of implications does this have for the amount of amplification required of our I/V converter?

We want to choose the amplification A of our I/V converter such that $I \cdot A$ lies within the input voltage range of the A/D converter. What can we now say about the accuracy of the measurement? When is it within 1 percent accurate?
The problem statement

We want to measure the resistance of a single atom ($R=12.9k\Omega$). We'll be using a voltage of $U=50mV$. A current-to-voltage converter (I/V-converter) is used to convert the measured current to a $10^x$ time higher voltage (1 Ampère is $10^x$ Volt). This voltage is then sent to a 16-bit A/D converter with an input range of $-5$ to $5V$. 
What should the value of $X$ be so that the resistance can be measured up to $1\%$ accuracy?
  
  
*
  
*(A) $X$ must be at least $3$; 
  
*(B) $X$ must be at least $4$; 
  
*(C) $X$ must be at least $4$ and maximally $6$ or 
  
*(D) $X$ must be at least $3$ and maximally $7$
  

I don't understand how we can distinguish the difference in accuracy between the cases. Any hints would be highly appreciated.

Our current is given by $I=\frac{U}{R}= 3.6nA$. The resolution of our A/D converter is given by $\frac{5--5}{2^{16}-1} = 1.5\cdot10^{-4} V$.  
Hence at an amplification of $10^3$ our signal would be picked up by our A/D converter, since this is within the resolution. At an amplification at 10^7 this is no longer the case, since $36V > 5V$. 
This leads me to go for (C); $X$ must be at least $4$ and maximally $6$. I do not know, however, how to calculate the accuracy in each case. How would you determine the accuracy?
 A: Two things are important to note when we want to discuss whether a
measurement lies within a certain accuracy. Can we measure the value
that we want to measure, i.e. does our measurement value lie within
the dynamic range? Do we have enough resolution to resolve a measurement up to a certain
percentage of our measurement value?
Well, since we know our measurement value --- it is
$I=\frac{50mV}{12.9k\Omega} = 3.6nA$, the first question we need
to ask is: how much do we need to amplify this signal to measure it
properly? 
This give rise to another question. Whether we can measure something
or not is determined by the resolution of our system. It is
$$\frac{U_{max}-U_{min}}{2^{bits}-1}=\frac{5V--5V}{2^{16}-1} = 1.5mV
\ .$$
Hence, our minimal amplication is given by $$\frac{0.15mV}{3.6nA} =
41.6 V/A \ .$$
Since our question is asking for an integer exponent $x$ of $10$, we need
$x>3$.

We now look into the accuracy requirement. If we want an accuracy of
$1\%$, this is equivalent to saying we want to be able to measure a
deviation of up to $1\%$ of our measurement value $R=12.9k\Omega$. Put
$\Delta R := 0.01 \cdot R = 129 \Omega$. In order to measure this we
need to measure a current of $\Delta I = \frac{50mV}{129\Omega} =
0.39mA$. This is accomplished even with the minimal amount of
amplification $10^3$, since this is solely determined by our ADC
resolution $0.15mV$ which is more that twice sufficient.
We conclude by remarking that at an amplification of $10^7$, our
measurement voltage value will be $36V$, which lies outside the range
of the input voltage of our ADC. Hence, we are left with the case of
(C), as the only viable option: $x$ must be at least $4$ and maximally
$6$.
