2. There are three consecutive even numbers. For the first one, we found the largest even proper divisor, for the second one - the largest odd proper divisor, and for the third one - again the largest even proper divisor. Can the sum of the three obtained divisors be equal to 2013? (A divisor of a natural number is called proper if it is different from 1 and the number itself)
Answer: Yes, it can. Solution. Here is an example: 1340, 1342, and 1344. For the first number, the largest even divisor is 670, for the third number, it is 672, and for the second number, the largest odd divisor is 671. $670+671+672=2013$. Remark. There are two natural ways to come up with this example. One can try to select a triplet so that the first number in it is divisible by 4, i.e., has the form $4 n$. Then the next number is $4 n+2$, and the third is $4 n+4$. But then it is clear that the divisors in question are $2 n, 2 n+1,2 n+2$. It remains to solve the equation $2 n+(2 n+1)+(2 n+2)=2013$. Or one can simply write the number 2013 as $670+671+672=2013$ and double each of the addends.