Suppose N is the page number of the highest-numbered missing page, and M is the page number of the last page before the missing pages.

The sum of all page numbers through page N is (N)(N+1)/2. The sum of all page numbers through page M is (M)(M+1)/2. The sum of the page numbers of the missing pages (S) is

S = (N)(N+1)/2 - (M)(M+1)/2

or

2S = N^2 + N - M^2 - M

= (N^2 - M^2) + (N - M)

= (N - M)(N + M) + (N - M)

= (N - M)(N + M + 1)

When we examine this product, we find it must be the product of an odd number and an even number. When we find the prime factorization of 2S = 2*9808 = 19616, we get

19616 = (2^5)(613)

Thus, we will only have an even and odd factor if we choose 613 as the odd factor and 32 as the even factor. We now have a "sum and difference problem" where the sum of our two numbers N and M must be 612, and their difference must be 32.

The two numbers N and M must be (612/2) ± (32/2) = {322, 290}.

The sum of all page numbers through page N is (N)(N+1)/2. The sum of all page numbers through page M is (M)(M+1)/2. The sum of the page numbers of the missing pages (S) is

S = (N)(N+1)/2 - (M)(M+1)/2

or

2S = N^2 + N - M^2 - M

= (N^2 - M^2) + (N - M)

= (N - M)(N + M) + (N - M)

= (N - M)(N + M + 1)

When we examine this product, we find it must be the product of an odd number and an even number. When we find the prime factorization of 2S = 2*9808 = 19616, we get

19616 = (2^5)(613)

Thus, we will only have an even and odd factor if we choose 613 as the odd factor and 32 as the even factor. We now have a "sum and difference problem" where the sum of our two numbers N and M must be 612, and their difference must be 32.

The two numbers N and M must be (612/2) ± (32/2) = {322, 290}.