This problem is a good example of how statistics can be used to solve exercises like this. Notice that this expression looks suspiciously like the formula for the weighted arithmetic mean, which is used to compute the arithmetic mean of grouped sequences. The formula I am talking about is the following:

$\bar{x}=\frac{f_1x_1+f_2x_2+...+f_kx_k}{N}$

Where $f_i$ are the frequencies of the data set, and $N$ is the size of the data set. If we treat the above expression as the arithmetic mean of a data set, we get the following data set.

We know that if a data set is symmetric, its mode (if it exists and if it's unique), its median and its arithmetic mean are equal. This data set is very close to being symmetric, so the mode and median will also be very close to the arithmetic mean. The median is quite hard to compute without a calculator, but we can easily see that the mode is $104$ , it having the highest frequency. So, the expression above - the arithmetic mean, is very close to this value, so we can conclude that it being rounded to the nearest digit will yield $104$ .

$\textcolor{#69047E}{\text{Let the nos. be a,b,c,d,e,f,g}}$ ( $\tiny{\text{pls don't judge me for being lazy}}$ )

$\dfrac{a.101+b.102+c.103+d.104+e.105+f.106+g.107}{a+b+c+d+e+f+g}$

101 can be written as 104-3 Similarly,

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Re-writing the expression :

$\boxed{\dfrac{104(a+b+c+d+e+f+g)-3a-2b-c+e+2f+3g}{a+b+c+d+e+f+g}}$

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Notice that

• a+2=g
• b-1=f
• c+3=e

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Re-writing it again :

$\boxed{\dfrac{104(2a+2b+2c+d+4)-7}{2a+2b+2c+d+4}}$

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$\textcolor{#3D99F6}{\text{Let 2a+2b+2c+d+4 be T }}$

$=>\dfrac{104(T)}{T} - \dfrac{7}{T}$

Wait ,don't you think $\dfrac{7}{T}$ is v.v. smol? Well yes it is!

So the nearest integer is $\boxed{\large{104}}$

Boom!

Let the quotient to be:

$Q = \frac {101a_1 + 102a_2 + 103a_3+104a_4+105a_5+106a_6+107a_7}{a_1 + a_2 + a_3+a_4+a_5+a_6+a_7}$

where $a_1 = 15793475837326$ , $a_2 = 25734263464529$ , $a_3 = 34957685947379$ , $a_4 = 67483926829160$ , $a_5 = 34957685947382$ , $a_6 = 25734263464528$ , and $a_7 = 15793475837328$ .

We note that $a_1 \approx a_7$ , $a_2 \approx a_6$ , and $a_3 \approx a_5$ , with an error of $< 2 \times 10^{-13}$ . Therefore,

$\begin{aligned} Q & \approx \frac {(101+107)a_1+(102+106)a_2+(103+105)a_3 + 104a_4}{2a_1 + 2a_2 + 2a_3 + a_4} \\ & = \frac {104(\cancel{2a_1 + 2a_2 + 2a_3 + a_4})}{\cancel{2a_1 + 2a_2 + 2a_3 + a_4}} \\ & = \boxed{104} \end{aligned}$