The "logic" is that we can use a mod distributive law to pull out a common factor $\,\color{#c00}{c=5},\,$ i.e.
$$ \bbox[1px,border:3px solid #c60]{\bbox[8px,border:1px solid #c00]{\color{#c00}ca\bmod \color{#c00}cn\:\! =\, \color{#c00}c(a\bmod n)}} $$
This decreases the modulus from $\,cn\,$ to $\,n, \,$ simplifying modular arithmetic. Also it may eliminate CRT = Chinese Remainder Theorem calculations, eliminating needless inverse computations, which are much more difficult than above for large numbers (or polynomials, e.g. see this answer).
This mod distributive law is often more convenient in congruence form, e.g.
$$ \ \ \bbox[1px,border:3px solid #c60]{\bbox[8px,border:1px solid #c00] {ca\equiv c(a\bmod n)\ \ \ {\rm if}\ \ \ \color{#0a0}{cn\equiv 0}\ \pmod{\! m}}}$$
since we have: $\,\ c(a\bmod n) \equiv c(a\! +\! kn)\equiv ca+k(\color{#0a0}{cn})\equiv ca\pmod{\!m}$
e.g. in the OP: $\ \ I\ge 1\,\Rightarrow\, 10^{\large I+N}\!\equiv 10^{\large I}(10^{\large N}\!\bmod 7)\ \ {\rm by} \ \ 10^I 7\equiv 0\pmod{\!35}$
Let's use that. Recall that exponents on $10$ can be reduced mod $\,6\,$ by little Fermat,
i.e. notice that $\ \color{#c00}{{\rm mod}\,\ 7}\!:\,\ 10^{\large 6}\equiv\, 1\,\Rightarrow\, \color{#c00}{10^{\large 6J}\equiv 1}.\, $ So if $\ I \ge 1\ $ then as above
$\ \phantom{{\rm mod}\,\ 35\!:\,\ }\color{#90f}{10^{\large I+6J}}\!\equiv 10^{\large I} 10^{\large 6J}\!\equiv 10^{\large I}(\color{#c00}{10^{\large 6J}\!\bmod 7})\equiv \color{#0a0}{10^{\large I}}\,\pmod{\!35} $
but $\ 5^{\large 102} = \color{#90f}{1\!+\!6J}\ $ by $\ {\rm mod}\,\ 6\!:\,\ 5^{\large 102}\!\equiv (-1)^{\large 102}\!\equiv 1$
So $\ 10^{\large 5^{\large 102}}\!\! = \color{#90f}{10^{\large 1+6J}}\!\equiv \color{#0a0}{10^{\large 1}} \pmod{\!35}\ $
Remark $\ $ For many more worked examples see the complete list of linked questions. Often this distributive law isn't invoked by name. Rather its trivial proof is repeated inline, e.g. from a recent answer, using $\,cn = 14^2\cdot\color{#c00}{25}\equiv 0\pmod{100}$
$\begin{align}&\color{#c00}{{\rm mod}\ \ 25}\!:\ \ \ 14\equiv 8^{\large 2}\Rightarrow\, 14^{\large 10}\equiv \overbrace{8^{\large 20}\equiv 1}^{\rm\large Euler\ \phi}\,\Rightarrow\, \color{#0a0}{14^{\large 10N}}\equiv\color{#c00}{\bf 1}\\[1em]
&{\rm mod}\ 100\!:\,\ 14^{\large 2+10N}\equiv 14^{\large 2}\, \color{#0a0}{14^{\large 10N}}\! \equiv 14^{\large 2}\!\! \underbrace{(\color{#c00}{{\bf 1} + 25k})}_{\large\color{#0a0}{14^{\Large 10N}}\!\bmod{\color{#c00}{25}}}\!\!\! \equiv 14^{\large 2} \equiv\, 96\end{align}$
This distributive law is actually equivalent to CRT as we sketch below, with $\,m,n\,$ coprime
$\begin{align} x&\equiv a\!\!\!\pmod{\! m}\\ \color{#c00}x&\equiv\color{#c00} b\!\!\!\pmod{\! n}\end{align}$
$\,\Rightarrow\, x\!-\!a\bmod mn\, =\, m\left[\dfrac{\color{#c00}x-a}m\bmod n\right] = m\left[\dfrac{\color{#c00}b-a}m\bmod n\right]$
which is exactly the same form solution given by Easy CRT. But the operational form of this law often makes it much more convenient to apply in computations versus the classical CRT formula.
Fractional extension of the mod distributive law: e.g. from here
notice: $\,\ \ \dfrac{\color{#c00}{11}}{35}\bmod \color{#c00}{11}(9)\,=\, \color{#c00}{11}(\color{#0a0}8)\,$ by $\color{#0a0}{\bmod 9\!:\ \dfrac{1}{35}\equiv \dfrac{1}{-1}\equiv 8},\ $ by
Theorem $\ \ \dfrac{\color{#c00}ab}d\bmod \color{#c00}ac\, =\, \color{#c00}a\left(\color{#0a0}{\dfrac{b}d\bmod c}\right)\ \ $ if $\ \ (d,ac) = 1$
${\bf Proof}{\bf \:\!_1}\ $ Bezout & $(d,ac)\!=\!1 \Rightarrow$ exists $\, d' \equiv d^{-1}\!\pmod{\!ac}.\,$ Factoring out $\,\color{#c00}a\,$ by mDL
$\qquad\qquad\ \ \ \ \dfrac{\color{#c00}ab}d\bmod \color{#c00}ac \,=\,\color{#c00}abd'\bmod \color{#c00}ac\, =\ \color{#c00}a(bd'\bmod c) = \color{#c00}{a}\left( \dfrac{b}d\bmod c\right)$
by inverses persist at modulus factors: $\ dd' \equiv 1\pmod{\!ac}\,\Rightarrow\, dd' \equiv 1\pmod{\!c}$
$\begin{align}{\bf Proof\:\!_2}\qquad\qquad
\color{#0a0}x \,&\:\color{#0a0}{\equiv b/d \pmod c}\\[.2em]
\iff\ \ \ dx&\equiv b\ \ \ \ \pmod{c},\ \ \ \ {\rm by}\ \ (d,c)=1\\[.2em]
\smash[t]{\overset{\times\ a}\iff}\ \ dax&\equiv ab\ \ \pmod{ac}\\[.2em]
\iff\ \ \ \color{#c00}a\color{#0a0}x &\equiv \color{#c00}ab/d\!\!\!\pmod{\color{#c00}ac},\ \ {\rm by}\ \ (d,ac)=1
\end{align}$
Best Answer
Here's a relatively fast method for general problems like these called square-and-multiply:
We can write $47 = 2^5 + 2^3 + 2^2 + 2^1 + 2^0. \ $ Therefore, $2^{47} = 2^{2^5} \cdot 2^{2^3} \cdot 2^{2^2} \cdot 2^{2} \cdot 2. \ $ And now we can compute these values $\big\{2^{2^k} \big\}_{k=0}^5$ via iterative squaring (and reducing modulo $899$ when necessary):
$\qquad \bullet \ \quad 2^2 \ = \ 4$
$\qquad \bullet \ \quad 2^{2^2} = \ 4^2 = \ 16$
$\qquad \bullet \ \quad 2^{2^3} \ = \ (16)^2 \ = \ 256$
$\qquad \bullet \ \quad 2^{2^4} \ = \ 256^2 \ = \ 65536 \ \equiv \ 808 \pmod{899}$
$\qquad \bullet \ \quad 2^{2^5} \ = \ 808^2 \ = \ 652864 \ \equiv \ 190 \pmod{899}$
Multiply the appropriate values together, reduce modulo $899$, and you're done(!!). Clearly, this is not the most computationally efficient approach in this specific case—see Bill Dubuque's answer (+1). But marvel at just how few computations there are relative to the naïve approach!
In general, if the exponent consists of $n$ bits when written in base-$2$, then—if I'm not mistaken—one needs to perform a maximum of $2n-3$ multiplications when using square-and-multiply (plus modulo reductions, if desired). Thus, the number of multiplications required grows logarithmically with the size of the exponent, whereas the number of multiplications required for naïve exponentiation grows linearly.