$1867k =\ldots 1992$, $k\in\mathbb{Z}^+$. Find the minimum value of k. The number $1867$ is multiplied by a positive integer $k$. The last four digits of the product are $1992$. Determine the minimum value of $k$.
$1867k =\ldots 1992$
 A: Hint $\ $ Let $\rm\, k = \ldots\!dcba\,$ have undetermined digits. Multiplying and comparing digits yields
$$\rm\begin{eqnarray} 1867\,(\ldots\!dcba)\ &= &\rm\ (7+6\cdot10+8\cdot 10^2 + 10^3)\,(a + b\cdot 10 + c\cdot 10^2+ d\cdot 10^3 + (\cdots)\, 10^4)\\
 & =&\rm\ 7a + (6a\!+\!7b)\, 10 + (8a\!+\!6b\!+\!7c)\, 10^2 + (a\!+\!8b\!+\!6c\!+\!7d)\,10^3+ (\cdots)\, 10^4\\
 & =&\rm\, \ldots\!1992\end{eqnarray}$$
$10^0$ digit: $\rm\ mod\ 10\!:\ 2\equiv 7a\:\Rightarrow\: a\equiv 6,\:$ so $\rm\:7a = 42,\:$ with carry $\color{#C00}4$
$10^1$ digit: $\rm\ mod\ 10\!:\ 9\equiv 6\cdot6 + 7b+\color{#C00}4\equiv 7b\:\Rightarrow\:b\equiv 7,\,$ so sum $= 89,$ with carry $= \color{#0A0}8$
$10^2$ digit: $\rm\ mod\ 10\!:\ 9 \equiv 8\cdot6\!+\!6\cdot 7\!+\!7c\!+\!\color{#0A0}8 \equiv 7c\!+\!8\,\Rightarrow\,7c\equiv 1\,\Rightarrow\,c\equiv 3,\,$ sum $119,\,$ carry $\color{brown}{11}$
$10^3$ digit: $\rm\ mod\ 10\!:\ 1 \equiv 6 + 8\cdot 7 + 6\cdot 3+ 7d + \color{brown}{11}\equiv 7d +1\,\Rightarrow\, 7d\equiv 0\,\Rightarrow\,d\equiv 0$
Therefore $\rm\:a + b\cdot 10 + c\cdot 10^2+ d\cdot 10^3 = 376.\:$ Indeed $\rm\:1867\cdot 376 = 701992.$
Remark $\ $ Equivalently, we may solve it iteratively, examining it mod $10, 10^2, 10^3, 10^4.\:$  This works because the above system of modular equations has nice triangular form: with carries $\rm\:c_i$
$$\rm \left[\begin{array}{cccc} 7 & 6 & 8 & 1 \\ 0 & 7 & 6 & 8 \\ 0 & 0 & 7 & 6 \\ 0 & 0 & 0 & 7 \end{array}\right]\left[\begin{array}{c} d\\ \rm c\\ \rm b\\ \rm a\end{array}\right] + \left[\begin{array}{c}\rm c_3 \\\rm c_2\\ \rm c_1\\ 0\end{array}\right] \equiv  \left[\begin{array}{c} 1\\9\\9\\2\end{array}\right]$$
The point of writing it out in this slight longer form is to explicitly exhibit this special structure - so to make clear the reason it works. This is perhaps better known in the analogous power series case, e.g. for computing inverses, Newton iteration, etc.
A: Hint:  you can find the digits of $k$ from the right.  7 times what ends in 2?
A: If you want to crack a nut with a sledgehammer then write this equation as:
$1867x - 10000y = 1992$
for positive integers $x,y$. Then we are left with a Diophantine equation. Solve this in the usual fashion and rejig the numbers to find the "smallest" solution for $x$ such that $y$ is positive.
A: Tune the digits one by one.
Step 1: $186[7] \to 199[2]$. Multiply $1867$ by $6$ gives the correct last digit [2].
Now you have $1867\times6=\cdots120[2]$.
Step 2: $12[02] \to 19[92]$. The tens digit should $+9$. Recall that $7\times7=\cdots9$.
Now you have $1867\times76=\cdots18[92$].
Step 3: $1[892] \to 1[992]$. The hundreds digit sould $+1$. $7\times3=\cdots1$.
So, $1892\times\color{green}{376}=\cdots[1992]$.
