## Monday, 24 October 2011

### Arithmetic-10 (XAT-2009)

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

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.