Problem:
Find the remainder when a number is divided by 19, given that the same number leaves a remainder of 21 when divided by 114.

Solution:
Let's call the number we're trying to find the remainder for "x". We know:

x ≡ 21 (mod 114)

This means that "x" leaves a remainder of 21 when divided by 114. In other words, we can write:

x = 114a + 21

where "a" is some integer. We need to find the remainder when "x" is divided by 19. We can write:

x ≡ r (mod 19)

where "r" is the remainder we're trying to find. We want to use the equation we derived earlier to substitute for "x" in this equation:

114a + 21 ≡ r (mod 19)

We want to simplify this equation so we can solve for "r". We can start by simplifying the left-hand side:

114a + 21 ≡ 9a + 3 (mod 19)

We can simplify further by noting that 114 ≡ 6 (mod 19) and 9 ≡ -10 (mod 19):

6a + 21 ≡ -10a + 3 (mod 19)

We can simplify again by adding 10a to both sides:

16a + 21 ≡ 3 (mod 19)

Now we can subtract 21 from both sides:

16a ≡ -18 ≡ 1 (mod 19)

Finally, we can multiply both sides by the inverse of 16 modulo 19, which is 12:

a ≡ 12 (mod 19)

This means that "a" is of the form "19n + 12" for some integer "n". We can substitute this into our equation for "x":

x = 114a + 21 = 114(19n + 12) + 21 = 2166n + 1365

Now we can find the remainder when "x" is divided by 19 by taking "x" modulo 19:

x ≡ 2166n + 1365 ≡ 6n + 6 ≡ 6(n + 1) (mod 19)

This means that the remainder when "x" is divided by 19 is 6 times some integer plus 6. We can check this by taking some values of "n":

- If n = 0, then x ≡ 6 (mod 19)
- If n = 1, then x ≡ 12 (mod 19)
- If n = 2, then x ≡ 18 (mod 19)
- If n = 3, then x ≡ 4 (mod 19)

We see that the remainder when "x" is divided by 19 can be any integer of the form 6k + 6, where k is an integer. Therefore, the correct answer is (C) 7.