To solve this problem, we need to find the natural numbers N for which the largest factor (excluding 1 and N) is exactly 21 times the smallest factor. Let's analyze the problem step by step.


1. Let's assume the smallest factor of N is x.
2. Since the largest factor (excluding 1 and N) is 21 times the smallest factor, the largest factor can be written as 21x.
3. Now, let's express N as the product of its prime factors. Let the prime factorization of N be p1^a1 * p2^a2 * p3^a3 * ... * pn^an.
4. Since x is the smallest factor of N, it must be equal to p1^b1 * p2^b2 * p3^b3 * ... * pn^bn, where b1 <= a1,="" b2=""><= a2,="" b3=""><= a3,="" ...,="" bn=""><=>
5. Similarly, 21x can be expressed as p1^c1 * p2^c2 * p3^c3 * ... * pn^cn, where c1 >= a1, c2 >= a2, c3 >= a3, ..., cn >= an.
6. From step 4 and 5, we can conclude that for each prime factor pi, bi <=>
7. Since x is the smallest factor, it must be equal to p1^a1, which means b1 must be equal to a1.
8. Similarly, 21x must be equal to p1^c1, which means c1 must be equal to a1.
9. From step 7 and 8, we can conclude that a1 = b1 = c1.
10. Therefore, for each prime factor pi, ai = bi = ci.
11. Now, we can express x and 21x as x = p1^a1 * p2^a2 * p3^a3 * ... * pn^an and 21x = p1^a1 * p2^a2 * p3^a3 * ... * pn^an.
12. Dividing the equation 21x = p1^a1 * p2^a2 * p3^a3 * ... * pn^an by x = p1^a1 * p2^a2 * p3^a3 * ... * pn^an, we get 21 = p1^a1 * p2^a2 * p3^a3 * ... * pn^an.
13. The prime factors of 21 are 3 and 7. Therefore, the prime factors of N must be either 3 or 7.
14. We can write N as N = 3^x * 7^y, where x and y are non-negative integers.
15. Since N is a natural number, x and y can take values from 0 onwards.
16. Therefore, the possible values of N are 3^0 * 7^0, 3^1 * 7^0, 3^2 * 7^0, ..., 3^x * 7^0, ..., 3^0 * 7^y, ..., 3^