Latest IBM Aptitude Question SOLUTION: 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?

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 is 20.

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 only three consecutive numbers case and not for four)


If we sum up all, we get a total of 20 possibilities, which is not in the answer options 1 to 4

9999 not included in following answer from SUJITHA therefore 21 possibilities not 20.

Let X be denoted by aaab or baaa.

Since X is a multiple of 9.

3a + b must be a multiple of 9.

There will be a total of 20 possible cases.