# Difficult Number System Solved QuestionAptitude Discussion

 Q. Let $X$ be a four-digit number with exactly three consecutive digits being same and is a multiple of 9. How many such $X$’s are possible?
 ✖ A. 12 ✖ B. 16 ✖ C. 19 ✔ D. 20

Solution:
Option(D) is correct

Let the four digit number be $'aaab'$ or $‘baaa’$

Since, the number has to be a multiple of 9, therefore $3a + b$ should be either $9, 18$ or $27$.

Case I: $3a + b = 9$
Possible cases are: $(1116, 6111, 2223, 3222, 3330, 9000)$

Case II: $3a + b = 18$
Possible cases are: $(3339, 9333, 4446, 6444, 5553, 3555, 6660)$

Case III: $3a + b = 27$
Possible cases are: $(6669, 9666, 8883, 3888, 7776, 6777, 9990)$

Hence total number of cases 20.