Arrange the numbers in the given rectangular blocks Consider the set $\{ (1,1), (1,2), \cdots , (1,10), (2,1), (2,2), \cdots , (10,10)\}$. The set contains all possible pairs (ordered) of numbers involving integers from $1$ to $10$. There are $100$ pairs in total. You're given $10$ rectangular blocks such that each block can contain $10$ pairs of numbers. You've to put the pairs in these boxes such that the first element of the first pair of each box is same as the last element of the last pair in the same box. Also, two consecutively placed pairs should be of the form $\{ (a,b), (b,c)\}$ where $a$ is not necessarily equal to $c$. A particular pair (out of all $100$ pairs) can be put in only one box. Show that the given arrangement is possible, that is, you can put the pairs in the boxes under the given constraints.
I tried out a few examples and it seems to work for the lower numbers (not for $n=2$ though).
 A: Generalised, you have a complete digraph $K_n$ augmented with the $n$ self-loops, and you want to partition the edges into $n$ closed trails (like cycles but permitting repeated vertices).
When $n$ is odd this can be done in a very symmetric way: let trail $i$ be the cycle $$i \to i \to i+1 \pmod n \to i+3 \pmod n \to \ldots \to i+\frac{(n-1)n}2 \pmod n$$
with step sizes $0,1,2,\ldots,n-1$ (or, indeed, any other permutation of those sizes).
With $n$ even each such trail finishes on $i + \frac n2$, so it's not closed. However, that hints at a possible solution strategy: remove the steps $0$ and $\frac n2$, splitting the trail into four parts:


*

*Step size 0

*Step sizes $1,2,\ldots,\frac n2 - 1$

*Step size $\frac n2$

*Step sizes $\frac n2 + 1, \frac n2 + 2, \ldots, n-1$
(Parts 4 are parts 2 in reverse). For $0 \le i < \frac n2$ combine parts 2 and 4 from $i$ with two step sizes $\frac n2$: $i \to i + \frac n2 \pmod n \to i$. For $\frac n2 \le i <n$ combine parts 2 and 4 from $i$ with two step sizes $0$. Then the only challenge is to find a suitable matching so that self-loop is covered. For $n=10$ this is straightforward enough:
0 5 0 1 3 6 0 6 3 1
1 6 1 2 4 7 1 7 4 2
2 7 2 3 5 8 2 8 5 3
3 8 3 4 6 9 3 9 6 4
4 9 4 5 7 0 4 0 7 5

5 6 6 8 1 1 5 1 8 6
6 7 7 9 2 2 6 2 9 7
7 8 8 0 3 3 7 3 0 8
8 9 9 1 4 4 8 4 1 9
9 0 0 2 5 5 9 5 2 0

