## N6

Vukasin, Dimitrije, Dusan, Stefan and Filip asked their professor to guess a three consecutive positive integer numbers after they had told him these (true) sentences:

Vukasin: "Sum of the digits of one of them is a prime number. Sum of the digits of some of the other two is an even perfect number ( $n$ is perfect if $\sigma(n)=2 n$ ). Sum of the digits of the remaining number is equal to the number of its positive divisors."

Dimitrije:"Each of these three numbers has no more than two digits 1 in its decimal representation."

Dusan:"If we add 11 to one of them, we obtain a square of an integer."

Stefan:"Each of them has exactly one prime divisor less then 10."

Filip:"The 3 numbers are square-free."

Their professor gave the correct answer. Which numbers did he say?


Solution. Let the middle number be $n$, so the numbers are $n-1, n$ and $n+1$. Since 4 does not divide any of them, $n \equiv 2(\bmod 4)$. Furthermore, neither 3,5 nor 7 divides $n$. Also $n+1+11 \equiv 2(\bmod 4)$ cannot be a square. Then 3 must divide $n-1$ or $n+1$. If $n-1+11$ is a square, then $3 \mid n+1$ which implies $3 \mid n+10$ (a square), so $9 \mid n+10$ hence $9 \mid n+1$, which is impossible. Thus must be $n+11=m^{2}$.

Further, 7 does not divide $n-1$, nor $n+1$, because $1+11 \equiv 5(\bmod 7)$ and $-1+11 \equiv 3(\bmod 7)$ are quadratic nonresidues modulo 7. This implies $5 \mid n-1$ or $5 \mid n+1$. Again, since $n+11$ is a square, it is impossible $5 \mid n-1$, hence $5 \mid n+1$ which implies $3 \mid n-1$. This yields $n \equiv 4(\bmod 10)$ hence $S(n+1)=S(n)+1=S(n-1)+2(S(n)$ is sum of the digits of $n)$. Since the three numbers are square-free, their numbers of positive divisors are powers of 2 . Thus, we have two even sums of digits - they must be $S(n-1)$ and $S(n+1)$, so $S(n)$ is prime. From $3 \mid n-1$, follows $S(n-1)$ is an even perfect number, and $S(n+1)=2^{p}$. Consequently $S(n)=2^{p}-1$ is a prime, so $p$ is a prime number. One easily verifies $p \neq 2$, so $\mathrm{p}$ is odd implying $3 \mid 2^{p}-2$. Then
$\sigma\left(2^{p}-2\right) \geq\left(2^{p}-2\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)=2\left(2^{p}-2\right)$. Since this number is perfect, $\frac{2^{p}-2}{6}$ must be one, i.e. $p=3$ and $S(n-1)=6, S(n)=7$ and $S(n+1)=8$.

Since 4 does not divide $n$, the 2 -digit ending of $n$ must be 14 or 34 . But $n=34$ is impossible, since $n+11=45$ is not a square. Hence, $n=10^{a}+10^{b}+14$ with $a \geq b \geq 2$. If $a \neq b$, then $n$ has three digits 1 in its decimal representation, which is impossible. Therefore $a=b$, and $n=2 \cdot 10^{a}+14$. Now, $2 \cdot 10^{a}+25=m^{2}$, hence $5 \mid m$, say $m=5 t$, and $(t-1)(t+1)=2^{a+1} 5^{a-2}$. Because $\operatorname{gcd}(t-1, t+1)=2$ there are three possibilities:

1) $t-1=2, t+1=2^{a} 5^{a-2}$, which implies $a=2, t=3$;
2) $t-1=2^{a}, t+1=2 \cdot 5^{a-2}$, so $2^{a}+2=2 \cdot 5^{a-2}$, which implies $a=3, t=9$;
3) $t-1=2 \cdot 5^{a-2}, t+1=2^{a}$, so $2 \cdot 5^{a-2}+2=2^{a}$, which implies $a=2, t=3$, same as case 1 ).

From the only two possibilities $(n-1, n, n+1)=(213,214,215)$ and $(n-1, n, n+1)=$ $(2013,2014,2015)$ the first one is not possible, because $S(215)=8$ and $\tau(215)=4$. By checking the conditions, we conclude that the latter is a solution, so the professor said the numbers: 2013, 2014, 2015.

