The numbers are $2013, 2014, 2015$. Condition 1 is satisfied. $2013$'s digits sum up to $6$ which is a perfect number. For $2014$ the sum of digits is $7$ which is prime. For $2015$ the sum of digits is 8 and it has 8 factors as $2015=5*13*31$. The second condition is also obviously satisfied. For the third condition, note that $2014+11=2025=45^2$. The fourth condition is also true. $2013=3*61*11, 2014=2*19*53, 2015=5*13*31$. The fifth condition is also satisfied as they are square free. I don't have proof for uniqueness currently. Some things I found: The numbers have to be of the form $4k+1,4k+2,4k+3$. The number of factors is always a power of 2 so one of them has the sum of digits a power of 2. Maybe we can show with this that the sum of digits are $6,7,8$ respectively, which narrows down the possibilities.

Full solution?

We have given the following conditions about three positive integers: (C1) The numbers are consecutive. (C2) The sum of the digits of one number is a prime. (C3) The sum of the digits of another number is an even perfect number. (C4) The sum of the digits of the third number equals the number of it’s positive divisors. (C5) None of the three numbers has more than two digits equal to 1. (C6) If 11 is added to one of the numbers, the sum is a perfect square. (C7) The three numbers has exactly one prime divisor less than 10. (C8) The three numbers are squarefree. The only triple of integers satisfying (C1)-(C8) is $(2013,2014,2015)$. Proof: Let $N$ be the smallest of the three numbers. Then by (C1) and (C8) we obtain $(1) \;\; N+k \equiv k+1 \pmod{4}, k \in \{0,1,2\}$. If one of the three numbers are divisible by 10, then this number has both 2 and 5 as prime divisors, contradicting (C7). Hence none of the three numbers has 0 as it last digit, which according to (C1) implies (2) the sum of the digits of the three numbers are consecutive integers. A positive squarefree positive integer with $k$ prime divisors has $2^k$ positive divisors. Hence by (C4) (3) The sum of digits of one of the numbers is a power of 2. According to (C3) the sum of the digits $S$ of one of the numbers $M$ is a even perfect number. Hence $S = 2^{p-1}(2^p - 1)$, where $p$ is a prime. Assume $p \neq 2$. According to (C2) and (2) there exists an odd prime $q$ s.t. $|S - q| \leq 2$, which combined with the fact that $S$ is even give us $(4) \;\; | \, 2^{p-1}(2^p - 1) - q \, | = 1$. According to Fermat's little theorem $p \mid 2^{p-1}(2^p - 1) - 1$, which by (4) implies $(5) \;\; q = 2^{p-1}(2^p - 1) + 1$. From (C4) and (C8) we obtain that there is a positive integer $k$ s.t the sum of the digits is $2^k$. Then by (2) we fund that $q - 2^k \in \{-1,2\}$, yielding (since $q$ is odd) $(6) \;\; q = 2^k - 1$. The combination of (4) and (5) give us $(7) \;\; 2^{p-1}(2^p - 1) + 2 = 2^k$. Now $p \geq 3$ and $k>p$ by (7), yielding $4 | 2$ by (7). This contradiction implies $p=2$ and $S=6$. Therefore by (2) the sum of digits of any of these three numbers range from $6-2=4=2^2$ to $6+2=8=2^3$. Hence by (3) (8) the sum of the digits of the three numbers are either 4,5,6 or 6,7,8 respectively According to (C2) $N + k = x^2 - 11, k=0,1,2$ $x^2 = N + k + 11$ $x^2 = k+2 \pmod{3} \;\;\; \Rightarrow \;\;\; k \neq 0$ $x^2 = k \pmod{4} \;\;\; \Rightarrow \;\;\; k \neq 2$ Consequently $k=1$, which means there is a positive integer $M$ s.t. $(9) \;\; N = M^2-12$. Assume the last digit of the three numbers are 4,5,6. Hence $N+2 \equiv 6 \pmod{3}$, yielding $(10) \;\; 3 \mid N+2$. The sum of digits of $N$ is 4, which means the last digit $n_0$ of $N$ is either 1 or 3 (since $N$ is odd by (1)). If $n_0=1$, then $M^2-12 \equiv 1 \pmod{10}$, i.e. $M^2 \equiv 3 \pmod{10}$, a contradition implying $n_0=3$. Then the last digit of $N+2$ is 5, meaning 5 is a (prime) divisor of $N+2$, which is impossible by (C7) and (10). This combined with (8) implies (11) the sum of the digits of $N$, $N+1$, $N+2$ are 6,7,8 respectively. We know that $9 | N-6$, which means $3|N$. Moreover $2 | N+1$ by (1). Consequently by (C7) $(12) \;\; p \mid N+2, p \in \{5,7\}$. Combining (9) and (12) we obtain $p \mid M^2-10$, yielding $p \neq 7$. Hence $p=5$, meaning the last digit of $N+2$ is 5 (since none of the three numbers have zero as last digit). Hence the sum of the other remaining digits of $N+2$ is 3 by (8). Thus by (C5) the nonzero digits of $N+2$ are 1,2 and 5 as the last digit. Thus there exists two distinct positive integers $u$ and $v$ s.t. $(13) \;\: N+2=10^u + 2 \cdot 10^v + 5$, which combined with (9) give us $(14) \;\; M^2 = 10^u + 2 \cdot 10^v + 15$. Then $M^2 \equiv 10^u + 3 \pmod{4}$, yielding $u=1$, which inserted in (14) result in $M^2 = 2 \cdot 10^v + 25$, i.e. $(15) \;\; (M - 5)(M + 5) = 2^{v+1} \cdot 5^v$. Now $GCD(M-5,M+5)=10$ by (15), which according to (15) implies $(16) \;\; {\textstyle (\frac{M \pm 5}{10}, \frac{M \mp 5}{10}) = (2^{v-1}, 5^{v-2})}$. Therefore by (16) $\;\; {\textstyle \frac{M \pm 5}{10} \;–\; \frac{M \mp 5}{10} = 2^{v-1}-5^{v-2}}$, i.e. $(17) \;\; 2^{v-1}-5^{v-2} = \pm 1$. Assume $2^{v-1}-5^{v-2}=1$. Then $2^{v-1} > 5^{v-2}$, yielding ${\textstyle (\frac{5}{2})^{v-2} < 2}$. Hence $v=2$, which means $N+2=215$, which prime factorization is $5 \cdot 43$. But the fact that $8=2^3$ means $N+2$ has 3 prime divisor by (C4) and (C8). Consequently $2^{v-1} \: – \: 5^{v-2}=-1$. Define $f(x)=2^{x-1} \: – \: 5^{x-2}$ where $x \geq 2$. Then $f(x) \:– \: f(x+1) = (2^{x-1} \: – \: 5^{x-2}) \:–\: (2^x \:–\: 5^{x-1}) = 2 (2 \cdot 5^{x-2} \: – \: 2^{x-2}) > 0$, i.e. $f(x) > f(x+1)$. This combined with the fact $f(3)=-1$ give us $v=3$. Hence by (14) $M^2 = 10^1 + 2 \cdot 10^3 + 15 = 2025 = 45^2$, which according to (6) means $N = M^2 \:–\: 12 = 2025 \:–\: 12 = 2013$. Consequently the conditions (C1)-(C8) are satisfied only if the three numbers are 2013, 2014 and 2015. The prime factorization of these three numbers are $3 \cdot 11 \cdot 61$, $2 \cdot 19 \cdot 53$ and $5 \cdot 13 \cdot 31$. Therefore this triple of integers satisfies (C1)-(C8). q.e.d.

So to get 10/10 we should write all this solution!!!??? I think this exercise whould be really difficult for JBMO!!!