Since $x^2 - (e + \pi)x + e \pi = (x - \pi)(x - e)$ has transcendental roots, we know that the coefficients are not both rational, and not even algebraic (see comment by José).

My question is, can we determine, by either adding or multiplying digit-wise (mod 10) the digits of $\pi$ and $e$, whether the resulting numbers are transcendental?

Suppose $p_n$ and $e_n$ are the $n$th digits of $\pi$ and $e$ respectively. Let $d_n = p_n + e_n \ (\mbox{mod } 10)$. The new number $t$ will have $d_n$ on the $n$th digit. We can similarly define $t$ with $d_n = p_n \cdot e_n \ (\mbox{mod } 10)$. Is $t$ transcendental?

More general version of the question:

Since it might be very difficult to determine whether two individual numbers are transcendental, here is a more general question:

If we add or multiply the digits of $\pi$ or $e$ digit-wise (mod $n$), where $n$ is a digit from 2-10, which of the resulting numbers are transcendental?

In the general case, define $d_n = p_n + e_n \ (\mbox{mod } k)$ where $k$ is an integer in $\{2,3,4,5,6,7,8,9,10\}$. In this case, we can use a choice function to determine $k$, or determine $k$ from $n$ in some suitable way. In the general case there are many possibilities for the different kinds of numbers that can be formed.

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    $\begingroup$ I bet they're all transcendental, and I bet I won't be able to collect on my bet. $\endgroup$ Jun 14, 2019 at 6:16
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    $\begingroup$ “Either” $\pi+e$ or $\pi e$ is transcendental whether the pope is catholic or I am the queen of England. What is the question? $\endgroup$ Jun 14, 2019 at 7:16
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    $\begingroup$ @FrancoisZiegler: I believe the idea is to create two new operations, call them $+_{10}$ and $\times_{10}$, which come from identifying an element of $\mathbb{R}$ as a sequence of elements of $\mathbb{Z}_{10}$, $x \mapsto (\dots, 0, 0, \dots, a_n, a_{n-1}, \dots, a_0, a_{-1}, \dots)$, via the decimal expansion, and adding or multiplying componentwise. The question then is whether $\pi +_{10} e$ or $\pi \times_{10} e$ are (identified with) transcendental numbers. $\endgroup$ Jun 14, 2019 at 7:25
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    $\begingroup$ Oooh! I see now. What about $\sqrt 2+_{10}\sqrt 3$, $\sqrt 2\times_{10}\sqrt 3$? $\endgroup$ Jun 14, 2019 at 8:29
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    $\begingroup$ I edited the question, so I hope it is more clear. This is similar to another question I asked (though I had since deleted my account). $\endgroup$
    – user141903
    Jun 15, 2019 at 12:29