Why do Carmichael numbers (appear to) frequently end in $1$? An integer $n$ is a Carmichael number if it is composite and satisfies $a^{n-1} \equiv 1(\mod{n})$ for all integers $a$ with $1<a<n$ and $\gcd(a,b)=1$.  From the relevant OEIS: 
$$561, 1105, 1729, 2465, 2821, 6601, 8911, 10585, 15841, 29341, 41041, 46657, 52633, 62745, 63973, 75361, 101101, 115921, 126217, 162401, 172081, 188461, 252601, 278545, 294409, 314821, 334153, 340561, 399001, 410041, 449065, 488881, 512461$$
Are the first $33$ Carmichael numbers.  If we consider that Carmichael numbers must end in $1, 3, 5, 7$, and $9$, then with no further consideration one might begin with the very rough heuristic that out of the first $k$ Carmichael numbers approximately $\frac15$ must end in $1$, for reasonably large $k$, if you expect equitable distribution for whatever reason. Yet we have $20$ Carmichael primes which end in $1$ out of the first $33$, which very much so does not suggest any equitable distribution. I would like some clarification as to why this is, hopefully in an at least partly analytic approach.
edit: a quick scroll through a much longer list of Carmichael numbers similarly suggests a much greater amount one Carmichael numbers which end in $1$ than would seem normal.  Additionally, they almost always seem to be consecutive; this might just be due to the fact that there are more of them.
 A: Thomas has answered a very similar problem in the following thread:
Why are Carmichael Numbers less common with an arithmetic progression
The fact that you're seeing this phenomena has nothing to do with the number 10 specifically.
Using Korselt's criterion -  $n$ is a Carmichael number iff $n$ is composite, square free, and for all prime divisors of $n$, we have $p − 1 ∣ n − 1$.
Suppose you're "trying to assemble" a Carmichael number of the form $n = p\cdot k + a$, for some prime number $p$. For every prime factor, $q|n$, we'll have $\ q-1|(p\cdot k+a-1)$. Therefore, for $q$ to be of the form $p\cdot k + 1$, we must have $a=1$.
That means, that if $a \ne 1$ then $n$ won't have prime factors of the form $p\cdot k+1$, making it a lot harder to assemble Carmichael numbers, because you have less primes to work with (especially for small $p$'s). 
Note that for every prime $p \ne 2$ that we choose, $n=1 \ mod\  p\ $ iff $\ n=1\ mod\ 2p$, because all Carmichael numbers are odd. Then specifically for 10, the group of Carmichael numbers of the form $10\cdot k + 1$ are the same as the group of Carmichael numbers of the form $5\cdot k + 1$
The following table shows the number (denoted as $M$) of Carmichael numbers $n < 10^{16}$, satisfying $n = 1\ mod\ p$, for different primes, out of the total $246683$ Carmichaels numbers $n < 10^{16}$.
+---------------------+----------+
| p        | M        | M/246683 |
+---------------------+----------+
|3         |245288    |99.43%    |
|5         |215713    |87.44%    |
|7         |168856    |68.45%    |
|11        |100071    |40.56%    |
|13        |77178     |31.28%    |
|17        |55109     |22.34%    |
|19        |36363     |14.74%    |
+---------------------+----------+

A: The main reason is that if a Carmichael number $n$ has a prime of the form $10k+1$ in it, then $10|n-1$, and so $n\equiv1\pmod{10}$. If we assume the distribution of primes in a factorization is average according to last digit, then as only one needs to be of this form, the feature becomes increasingly likely (P($n$ is Carmichael $\land$ $n\equiv 1\pmod{10})\approx1-\left(\frac34\right)^k$ if $n$ has $k$ prime factors).
There is a list of factorizations of small 3-Carmichael numbers here, and the only other possibility is that the last digits product to $1\pmod{10}$, for example $23\times199\times353=1,615,681$ and $29\times113\times1093=3,581,761$.
