Proof for a modular arithmetic identity. How can I prove the following? Or is there a reference to the proof for this?
$$d \mod abc = d \mod a + a\left(\left\lfloor\frac{d}{a}\right\rfloor \mod b\right)+ab\left(\left\lfloor\frac{d}{ab}\right\rfloor \mod c\right)$$
 A: Let $d = r + ka; 0\le r < a$.  Then $r =$ what you are referring to as $d\mod a$.  And $k = [\frac da]$.
Let $k = s + mb; 0 \le s < b$ so $s = [\frac da] \mod b$ and $m =[\frac d{ab}]$
Let $m = t + nc; 0 \le t< c$ so $t = [\frac d{ab}] \mod c$
$d = r + ka=$
$r + (s+mb)a = r+(s+(t+nc)b)a = r + sa + tba + n*abc$
And $r + sa + tba \le  (a-1) + (b-1)a+ (c-1)ba = a-1 + ab -a +abc - ba = abc - 1$.
So $r + sa + tba = d \mod abc$.
And $r + sa + tba = d\mod a + a([\frac da] \mod b) + ab([\frac d{ab}] \mod c)$.
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We can try to do it more directly.  By division algorithm:
for and $n,q$ there are unique integers $d,r$ so that $n = dq + r$ were $0 \le r < q$.
It's simple to see that $dq \le n < dq + q$ so $d \le \frac nq < d+1$ so $d = [\frac nq]$.
And this exercise, although it makes my teeth hurt to see it, seems to be taking $n \mod q$ to be and operation that is defined to be equal to $r$ as a definition. (So be it, I'm tired of arguing).
So $n = (n\mod q) + q*[\frac nq]$ is an identity. Furthermore if $n = a + q*[\frac nq]$ it follows that $a = (n\mod q)$.
Hence  $d = (d \mod a) + a([\frac da])$
$= (d \mod a) + a([\frac da]\mod b + b([\frac {[\frac da]}b]))$
$=(d \mod a) + a([\frac da]\mod b + b([\frac {[\frac da]}b]\mod c + c([\frac{[\frac {[\frac da]}b]}c])))$
$= (d\mod a) + a([\frac da]\mod b) + ab([\frac {[\frac da]}b]\mod c) + abc*([\frac{[\frac {[\frac da]}b]}c])$.
So it only remains to show that $[\frac {[\frac da]}b] = [\frac a{ab}]$ and that $[\frac{[\frac {[\frac da]}b]}c] =[\frac a{abc}]$; something I glided over in my above proof.
Simple enough:
Lemma:  $[\frac {[\frac mn]}k] = [\frac m{nk}]$ (for positive integers).
Proof: $[\frac mn] \le \frac mn < [\frac mn]+1$
$\frac{[\frac mn]}k \le \frac m{nk} <  \frac{[\frac mn]}k + \frac 1k$ 
$[\frac{[\frac mn]}k] \le\frac{[\frac mn]}k \le \frac m{nk} <  \frac{[\frac mn]}k + \frac 1k$ 
Meanwhile $[\frac{[\frac mn]}k]+ 1= [\frac{[\frac mn]}k]+ \frac kk > \frac{[\frac mn]}k$.  And as $[\frac mn]$ is an integer.  $[\frac{[\frac mn]}k]+ \frac {k-1}k \ge \frac{[\frac mn]}k$.
So $[\frac{[\frac mn]}k]  \le \frac m{nk} <  \frac{[\frac mn]}k + \frac 1k\le [\frac{[\frac mn]}k]+ \frac {k-1}k +\frac 1k = [\frac{[\frac mn]}k]+1$
So $[\frac m{nk}]=[\frac{[\frac mn]}k]$
So 
$d= (d\mod a) + a([\frac da]\mod b) + ab([\frac {[\frac da]}b]\mod c) + abc*([\frac{[\frac {[\frac da]}b]}c])$
$= (d \mod a) +a([\frac da]\mod b) + ab([\frac {d}{ab}]\mod c)+ abc([\frac {d}{abc}])$.
And $d = (d\mod abc) + abc([\frac {d}{abc}])$.
So $d\mod abc = (d \mod a) +a([\frac da]\mod b) + ab([\frac {d}{ab}]\mod c)$
A: Numerical example:
$$\begin{align}95 \mod \color{red}2\cdot \color{green}3\cdot \color{blue}7 = \color{purple}{\mathbf{11}} \Rightarrow \frac{95}{\color{red}2\cdot \color{green}3\cdot \color{blue}7}&=\frac{(1+94)}{\color{red}2\cdot 3\cdot 7}=\\
&=\frac{1}{2\cdot 3\cdot 7}+\frac{2\cdot 47}{2\cdot 3\cdot 7}\\
&=\frac{1}{2\cdot 3\cdot 7}+\frac{2\cdot (2+45)}{2\cdot \color{green}3\cdot 7}=\\
&=\frac{1}{2\cdot 3\cdot 7}+\frac{2\cdot 2}{2\cdot 3\cdot 7}+\frac{2\cdot 3\cdot 15}{2\cdot 3\cdot 7}=\\
&=\frac{1}{2\cdot 3\cdot 7}+\frac{2\cdot 2}{2\cdot 3\cdot 7}+\frac{2\cdot 3\cdot (1+14)}{2\cdot 3\cdot \color{blue}7}=\\
&=\frac{1}{2\cdot 3\cdot 7}+\frac{\color{red}2\cdot 2}{2\cdot 3\cdot 7}+\frac{\color{red}2\cdot \color{green}3\cdot 1}{2\cdot 3\cdot 7}+2=\\
&=\frac{\color{purple}{\mathbf{11}}}{\color{red}2\cdot \color{green}3\cdot \color{blue}7}+2.\end{align}$$
