What are the numbers that leave a remainder of 5 when dividing 2007?
If the division of 2007 by some number leaves a remainder of 5, then this number divides \(2007-5=2002\). Thus, we calculate all the divisors of \(2002 = 2 \times 7 \times 11 \times 13\), listed in the right column of the following table.

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-065.jpg?height=340&width=854&top_left_y=895&top_left_x=632)

Since the remainder 5 must be less than the divisor, dividing 2007 by any of the following 14 numbers leaves a remainder of 5:

$$
7,11,13,14,22,26,77,91,143,154,182,286,1001 \text { and } 2002
$$