Given:

- Numbers 7179 and 9699 when divided by N leave the same remainder.
- We need to find the number of values of N that end with one or more than one zeroes.

Solution:

Let's first find the common remainder when 7179 and 9699 are divided by N.

Let the remainder be R.

So, we have:

- 7179 = xN + R
- 9699 = yN + R

where x and y are some natural numbers.

Subtracting the above equations, we get:

- 9699 - 7179 = (y - x)N
- 2520 = (y - x)N

Since 2520 is divisible by all numbers up to 10, we can say that N must also be divisible by all numbers up to 10.

Now, let's consider the second part of the question - finding the number of values of N that end with one or more than one zeroes.

- To end with one zero, N must be divisible by 10.
- To end with two or more zeroes, N must be divisible by 100 or more.

We can express N as a product of its prime factors:

- N = p1^a1 * p2^a2 * p3^a3 * ...

where p1, p2, p3, ... are prime numbers and a1, a2, a3, ... are their respective powers.

- To end with one zero, N must contain at least one factor of 2 and one factor of 5.
- To end with two or more zeroes, N must contain at least two factors of 2 and two factors of 5.

Let's count the number of such values of N.

- For N to contain at least one factor of 2 and one factor of 5, we have the following possibilities:

- p1 = 2, a1 >= 1 and p2 = 5, a2 >= 1
- p1 = 5, a1 >= 1 and p2 = 2, a2 >= 1
- p1 = 2, a1 >= 1 and p2 = 2, a2 >= 1 and p3 = 5, a3 >= 1
- p1 = 5, a1 >= 1 and p2 = 5, a2 >= 1 and p3 = 2, a3 >= 1

The number of possibilities for each case are:

- For p1 = 2 and p2 = 5: (1 + 1 + 1) * (1 + 1) - 1 = 5
(Here, we subtract 1 because we don't want to count the case where N = 1)
- For p1 = 5 and p2 = 2: (1 + 1 + 1) * (1 + 1) - 1 = 5
- For p1 = 2, p2 = 2 and p3 = 5: (1 + 1) * (1 + 1) * (1 + 1) - 1 = 7
- For p1 = 5, p2 = 5 and p3 = 2: (1 + 1) * (1