7.3. A natural number is called a palindrome if it remains unchanged when its digits are written in reverse order (for example, the numbers 4, 55, 626 are palindromes, while 20, 201, 2016 are not). Represent the number 2016 as a product of three palindromes greater than 1 (find all possible options and explain why there are no others).
Answer: 2-4.252. Solution. The number 2016 is not divisible by 11. Therefore, none of the palindromes are two-digit (they all are divisible by 11). This means that two of the palindromes are one-digit (if the two factors have no less than 3 digits, then the product is no less than $100 \cdot 100 > 2016$). The product of one-digit numbers is no less than 4, so the third factor T is no more than $2016 / 4 = 504$, that is, it is a three-digit number. We will consider 2 cases. For this, we need the prime factorization of 2016: $2016 = 2^5 \cdot 3^2 \cdot 7$.

Case 1. T < 200. Then T starts and ends with 1, that is, it is odd. However, the product of all odd primes in the factorization of 2016 is only 63, so all odd divisors of the number 2016 are not three-digit. This case is impossible.

Case 2. T $\geq 200$. Then the product of the other two factors is no more than 2016/200, that is, no more than 10. Only 6, 8, and 9 can be the product of one-digit divisors of the number 2016: $6 = 2 \cdot 3, 8 = 2 \cdot 4$, $9 = 3 \cdot 3$. The second option fits, since $2016 / 8 = 252$ is a palindrome; but the others do not, as $2016 / 6 = 336$ and $2016 / 9 = 224$ are not palindromes.

Criterion. Correct answer without justification - only 1 point.