The multiples of 11 will coincide whenever (K^2) – (4 x 5) is a multiple of 11. This occurs when K is either 3 more or 3 less than a multiple of 11, so 3, 8, 14, 19, 25, 30 etc.

The first three times this is a prime number are 3, 19 and 41.

Therefore the answer is 41.

For an explanation of why this is the case, read on:

Under the assumption that KM+5L and KL+4M are both congruent to 0, modulo 11, we can multiply either side by an integer and that congruence will be preserved. Therefore we can multiply the first expression by K (which is defined as an integer) and the second by 5. Therefore K^2.M+5LK and 5LK+20M are also both 0 mod 11.

Subtracting one expression from the other unifies the two expressions but maintains the divisibility by 11. Doing so makes the 5LK terms cancel out so we end up with (K^2 – 20)M is 0 mod 11. Therefore for our purposes, where we are interested in the special values of K, this occurs when K^2 – 20 is a multiple of 11, which is where we came in.

As a side note, if you’re wondering what happens if you follow the other strand and assume M is a multiple of 11, it’s not difficult to show that that means that L must also be a multiple of 11 for either/both of the original expressions to be a multiple of 11, which it will be for all K.