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)
21

(E)
None of the above

Answer follows here:

Solution:

“Sum of all digits must be divisible by 9” - for a
number to be multiple of 9.

The way to proceed is - considering three consecutive same digits and for the selection of fourth digit, checking divisibility by ‘9’ for
the complete number.

Let us see case by case:

Three consecutive 0’s: 9000 –
1 case

(Note here that as the sum of Three 0’s is ‘0’, the
fourth digit must be ‘9’)

Three consecutive 1’s: 1116, 6111 – 2 cases

(Note here that as the sum of Three 1’s is ‘3’, the
fourth digit must be ‘6’)

Three consecutive 2’s: 2223, 3222 – 2 cases

Three consecutive 3’s: 3330, 3339, 9333 –
3 cases

(Note here that as the sum of Three 3’s is ‘9’, the
fourth digit can be ‘0’ or ‘9’)

Three consecutive 4’s: 4446, 6444 – 2 cases

Three consecutive 5’s: 5553, 3555 – 2 cases

Three consecutive 6’s: 6660, 6669, 9666 –
3 cases

Three consecutive 7’s: 7776, 6777 – 2 cases

Three consecutive 8’s: 8883, 3888 – 2 cases

Three consecutive 9’s: 9990 – 1 case

(Note
that 9999 won’t be considered as it is asked for **consecutive numbers case and not for four)**__only three__
If we sum up all, we get a total of

**20**possibilities, which is not in the answer options 1 to 4.**Answer (5)**

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