aime26-12

Proposed solution

To solve the problem, we start by analyzing the given sum \( S_r \):

\[
S_r = \sum_{m \ge 0} \binom{10000}{502n + r}
\]

where \( \binom{10000}{k} = 0 \) if \( k > 10000 \). We need to find the number of integers in the list \( S_0, S_1, \ldots, S_{501} \) that are multiples of the prime number 503.

First, we recognize that \( S_r \) is a sum of binomial coefficients where the indices are congruent to \( r \) modulo 502. We can rewrite the binomial coefficient sum as:

\[
S_r = \sum_{n \ge 0} \binom{10000}{502n + r}
\]

We are interested in the values of \( S_r \) modulo 503. To do this, we use properties of binomial coefficients and the fact that 503 is a prime number. By Lucas' Theorem, we know that:

\[
\binom{10000}{k} \mod 503
\]

depends on the base-503 representation of 10000 and \( k \). We need to find the representation of 10000 in base 503. We start by calculating:

\[
10000 \div 503 \approx 19.857 \quad \Rightarrow \quad 10000 = 19 \times 503 + 437
\]

Thus, in base 503, 10000 is represented as \( 19 \cdot 503 + 437 \). Next, we represent 437 in base 503:

\[
437 \div 503 = 0 \quad \Rightarrow \quad 437 = 0 \cdot 503 + 437
\]

So, the representation of 10000 in base 503 is \( (19, 437) \).

For \( S_r \), we need to consider the binomial coefficients \( \binom{10000}{502n + r} \). We know that \( 502 = 503 - 1 \), so we need to analyze the binomial coefficients modulo 503. The binomial coefficient \( \binom{10000}{502n + r} \) will be non-zero only if \( 502n + r \le 10000 \). This implies \( n \) must be such that \( 502n + r \le 10000 \).

We can express \( 10000 \) in base 503 as \( (19, 437) \). The binomial coefficient \( \binom{10000}{k} \) modulo 503 can be evaluated using the base-503 representation. For \( \binom{10000}{502n + r} \) to be non-zero, \( 502n + r \) must be less than or equal to 10000. This means \( n \) must be such that \( 502n + r \le 10000 \).

Next, we consider the sum \( S_r \) modulo 503. We need to find the values of \( S_r \) for \( r = 0, 1, 2, \ldots, 501 \). We know that:

\[
S_r = \sum_{n \ge 0} \binom{10000}{502n + r}
\]

We need to determine when \( S_r \) is a multiple of 503. We use the fact that the binomial coefficients \( \binom{10000}{502n + r} \) are non-zero only if \( 502n + r \le 10000 \). We need to find the values of \( r \) for which \( S_r \equiv 0 \mod 503 \).

Since 503 is a prime number, we can use properties of binomial coefficients modulo a prime. We know that the binomial coefficients \( \binom{10000}{k} \) modulo 503 will be non-zero for \( k \) in the range from 0 to 10000. We need to check the values of \( S_r \) for \( r = 0, 1, 2, \ldots, 501 \) and see when they are multiples of 503.

We can use the fact that the binomial coefficients \( \binom{10000}{k} \) modulo 503 are symmetric around \( k = 5000 \) (since 10000 is even). This symmetry implies that the binomial coefficients \( \binom{10000}{k} \) and \( \binom{10000}{10000 - k} \) are congruent modulo 503. Therefore, the sums \( S_r \) for \( r = 0, 1, 2, \ldots, 501 \) will exhibit a certain symmetry.

To find the number of \( S_r \) that are multiples of 503, we need to check the values of \( S_r \) modulo 503. We know that \( S_r \) is a sum of binomial coefficients, and we need to determine when this sum is congruent to 0 modulo 503. We can use the fact that the binomial coefficients \( \binom{10000}{k} \) modulo 503 are symmetric and periodic.

We can use the fact that the binomial coefficients \( \binom{10000}{k} \) modulo 503 are periodic with period 503. Therefore, we need to check the values of \( S_r \) for \( r = 0, 1, 2, \ldots, 502 \) (since the binomial coefficients repeat every 503). We need to find the number of \( S_r \) that are multiples of 503.

We can use the fact that the binomial coefficients \( \binom{10000}{k} \) modulo 503 are symmetric around \( k = 5000 \). Therefore, the sums \( S_r \) for \( r = 0, 1, 2, \ldots, 502 \) will exhibit a certain symmetry. We need to chec