How many numbers lie between 11 and 1111 which when divided by 9 leave

lcm = 63
a = 33

a+(n-1) d < 1111

giving n = 18

N = 9K + 6 = 6,15,24,33,

N = 21m + 12 = 12, 33..

=> N = 63n + 33

Max value of N < 1111 = 63* 17 + 33

So total # = 18

Because n = 0 is the first number's index.

Answer = C

Check this link for more on how to solve such questions:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... s-part-ii/

The first number that leaves a remainder of 6 when divided by 9 is 6.

Then we have 6, 15, 24, 33, 42, 51, 60, and so on.

The first number that leaves a remainder of 12 when divided by 21 is 12.

Then we have 12, 33, 54 and so on.

We can see that 33 is the smallest number that satisfies both conditions.

Since the LCM of 9 and 21 is 63, we can keep adding 63 to 33 to find the succeeding numbers that satisfy both properties:

96, 159, ...

While we could list the numbers until we reach but not exceed 1111, a more efficient way is to find the find the largest number less than 1111 that satisfies both conditions, by the following equation:

33 + 63n < 1111

63n < 1078

n < 17.11

Since n is an integer, n = 17. That is, 33 + 63(17) is the largest integer that satisfies both conditions. Since 33 + 63(0) = 33 is the smallest integer that satisfies both conditions, we have 18 integers (in the form of 33 + 63m, where m is an integer from 0 to 17, inclusive) that satisfies both conditions.

Answer: C

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