Task 3. Determine three consecutive natural numbers whose product is 24 times greater than the greatest common divisor of the numbers 455 and 1001.
Solution. The numbers 455 and 1001 factored are $455=5 \cdot 7 \cdot 13$ and $1001=7 \cdot 11 \cdot 13$. The greatest common divisor of 455 and 1001 is $7 \cdot 13=91$. From the condition of the problem, the product of three consecutive numbers is equal to 91$\cdot$24=2184. Since $2184=2 \cdot 2 \cdot 2 \cdot 3 \cdot 7 \cdot 13$, the required three consecutive numbers are $12,13,14$.