Given: A number leaves a remainder of 79 when divided by 837.

To find: The remainder when the same number is divided by 31.

Approach:

Let's assume the number to be x.

When x is divided by 837, it leaves a remainder of 79, which can be written as:

x = 837a + 79, where a is a quotient.

We need to find the remainder when x is divided by 31.

We can use the Chinese Remainder Theorem to solve this problem.

Chinese Remainder Theorem:

If we have a system of congruences of the form:

x ≡ a1 (mod m1)
x ≡ a2 (mod m2)
.
.
.
x ≡ ak (mod mk)

where m1, m2, ..., mk are pairwise relatively prime integers, then the system of congruences has a unique solution modulo M = m1 * m2 * ... * mk.

The solution is given by:

x ≡ (a1 * M1 * N1) + (a2 * M2 * N2) + ... + (ak * Mk * Nk) (mod M)

where M1 = M/m1, M2 = M/m2, ..., Mk = M/mk, and N1, N2, ..., Nk are the modular multiplicative inverses of M1, M2, ..., Mk modulo m1, m2, ..., mk respectively.

Solution:

In this case, we have:

x ≡ 79 (mod 837)
x ≡ ? (mod 31)

837 and 31 are not relatively prime, so we cannot directly use the Chinese Remainder Theorem.

However, we can use the fact that:

x ≡ 79 (mod 837)
837 ≡ 6 (mod 31)

to simplify the problem.

We can write:

x ≡ 79 (mod 837)
x ≡ 79 (mod 31) [Since 837 ≡ 6 ≡ 0 (mod 31)]

Now the moduli are relatively prime, so we can use the Chinese Remainder Theorem.

M = 837 * 31 = 25947

M1 = M/837 = 31
M2 = M/31 = 837

31 * N1 ≡ 1 (mod 837)
N1 = 208

837 * N2 ≡ 1 (mod 31)
N2 = 13

Therefore, the solution is:

x ≡ (79 * 31 * 208) + (79 * 837 * 13) (mod 25947)
x ≡ 24673 (mod 25947)

The remainder when x is divided by 31 is:

24673 mod 31 = 17

Therefore, the answer is (D) 17.