Introduction

The problem at hand involves determining the remainder when a large sum of numbers, represented as repeated sequences of digits, is divided by 37. This problem involves an intricate interplay of number theory, specifically modular arithmetic and properties of prime numbers. In this article, we will walk through the steps to solve this, using both theory and practical application.

Step-by-Step Solution

Step 1: Representing the Numbers

The numbers in question are formed by repeating digits from 1 to 7, n times. For example, 111 for n3 or 222 for n2. Each of these can be mathematically represented as follows:

111... n times 1×(10^n - 1)/9 222... n times 2×(10^n - 1)/9 333... n times 3×(10^n - 1)/9 ...777... n times 7×(10^n - 1)/9

Step 2: Summing the Numbers

To find the total sum S, we sum the series above:

S 1×(10^n - 1)/9 2×(10^n - 1)/9 3×(10^n - 1)/9 ... 7×(10^n - 1)/9

Factoring out the common term gives:

S (1 2 3 4 5 6 7)×(10^n - 1)/9

The sum of the digits 1 to 7 is 28, so:

S 28×(10^n - 1)/9

Step 3: Finding the Remainder when Divided by 37

Our goal is to find the remainder when S is divided by 37:

S 28×(10^n - 1)/9

Step 4: Calculating 10^n mod 37

To simplify this, we use Fermat's Little Theorem, which states that for any integer a and a prime p, a^p ≡ a (mod p). For 37 (a prime number), we have:

10^36 ≡ 1 (mod 37)

This implies that 10^n mod 37 will repeat every 36 numbers. We need to consider n mod 36, which gives us the remainder r when n is divided by 36:

10^n ≡ 10^r (mod 37)

Step 5: Finding 10^n - 1 mod 37

Thus, we have:

10^n - 1 ≡ 10^r - 1 (mod 37)

Step 6: Substitute Back into the Sum

Substituting this back into the sum S, we get:

S ≡ 28×(10^r - 1)/9 (mod 37)

Step 7: Finding 28×10^r - 1 mod 333

To compute the final result, we need the modular inverse of 9 modulo 37. Using the Extended Euclidean Algorithm:

37 4×9 1

Therefore, 1 37 - 4×9, so 9^-1 ≡ -4 ≡ 33 (mod 37)

This means:

S ≡ 28×10^r - 1 × 33 (mod 37)

Step 8: Final Calculation

Finally, we can compute this for specific values of r from 0 to 35. For large n, r will take on values from 0 to 35, so the average case will generally yield:

S ≡ 0 (mod 37)

Conclusion

In summary, the remainder when the large sum S is divided by 37, when n is sufficiently large, is typically 0. This result emerges from the cyclic nature of 10^n mod 37 and the averaging out of contributions across the range of r values.