Buzzkill...

The eighth integer, denoted by $x$ , satisfies that $|x|^x$ is undefined. The exponentiation to any nonzero integer is defined, so $x = 0$ (otherwise the exponentiation is defined, contradicting the eighth integer condition). Hence the product of all integers should be $0$ . It just remains to check that none of the other integers produce a contradiction. (If there's a contradiction, the product is thus undefined.)

We interpret " $a$ is five times larger than $b$ " as $a = 5b$ , since otherwise the second integer is undefined (it is five times larger than $0$ , which should be $0$ again, but if we take "larger" to be strictly larger, then $0 > 0$ , contradiction). Similar interpretation applies to the third integer.

The fourth integer is ambiguous ( $0$ or $1$ ), but the product remains the same, so this is of no concern. We're looking for contradictions, not ambiguity.

The tenth integer is safe: it's defined as the number of positive integers less than the argument, so whenever $k \le 0$ , we have $\varphi(k) = 0$ . The problem is only when the argument is not an integer; in this case, "divisors of that number" is undefined. Luckily as $\varphi$ counts the number of something, its value has to be a positive integer, and also $12^{12}$ is an integer, so $\varphi(\varphi(12^{12}))$ is defined.

The eleventh integer might pose a problem if the sum is $0$ (it has infinitely many divisors). So we check the second, fourth, and fifth integer. The second integer is $0$ if we take that the answer to this problem is not undefined. The fourth integer can be either $0$ or $1$ . The fifth integer is $-20$ . The sum is thus $-20$ or $-19$ , both safe (only finitely many divisors, so the number of divisors can be determined, although it's unclear whether to take the number of positive divisors or all divisors).

The twelfth integer's description is quite unnatural: product of (number of prime numbers less than the product of the seventh integer and Ramanujan's number) is simply the parenthesized value (because the product of a single integer is itself). But this number is still defined.

The sixteenth integer might pose a problem if there are less than $16$ palindromes that are less than $9 \cdot 10^{12}$ . Luckily, there are more than $16$ : we simply observe that $1,2,3,\ldots,9,11,22,33,\ldots,99$ are palindromes, and there are $18$ of them, so the 16th smallest palindrome will be defined.

All the remaining integers are trivially defined, although perhaps not uniquely. So the integers are defined, and hence the product, and also the last three digits of the product, is $\boxed{0}$ .

(Observe that "undefined" is also a valid answer to the problem, since the second integer will hence also be undefined, and the product of somethings with undefined is also undefined.)

When I saw that this problem had this huge paragraph, I looked down at the bottom sentence and saw that it was looking for the product of the sixteen integers, so I (mentally) exclaimed "This ... better be $0$ or it's the most trollish question I've ever seen!" So I put $0.$

The $0$ comes from the $8\text{th}$ integer. $\left|0\right|^0$ is undefined because $0^0$ is undefined.

The only important information here is the eighth integer, we need an integer $x$ such that $| x | ^x$ is impossible to compute, so $x = 0$ only. Thus the eighth integer must be $0$ and hence the answer is $\boxed{0}$

This is probably the best question ever. The fifth integer is 0 as there are no known perfect numbers at the moment. Therefore the product of all the integers is 0.