Arrangements and Divisibility

IMO D

shankar.ashwin wrote:Different four-digit numbers are formed using the digits 1, 2, 3, 4, 5 and 6 (repetition is allowed). If 'n' numbers out of these are divisible by 3, then the value of 'n' is:

A) 234
B) 242
C) 324
D) 432
E) 464

To yield a multiple of 3, the sum of the 4 digits must be a multiple of 3.

First 3 digits:
The first 3 digits selected can be any of the 6 digits.
Number of options for each digit = 6.
Total options = 6*6*6 = 216.

Last digit:
The last digit selected must yield a sum that is a multiple of 3.
If the sum of the first 3 digits is a multiple of 3, then the last digit selected must be 3 or 6, so that the sum of all 4 digits is a multiple of 3.
If the sum of the first 3 digits is 1 more than a multiple of 3, then the last digit selected must be 2 or 5, so that the sum of all 4 digits is a multiple of 3.
If the sum of the first 3 digits is 2 more than a multiple of 3, then the last digit selected must be 1 or 4, so that the sum of all 4 digits is a multiple of 3.
Thus, whatever the sum of the first three digits, in each case the number of options for the last digit = 2.

To combine our options for the first 3 digits and our options for the last digit, we multiply:
216*2 = 432.

The correct answer is D.

using above digits, in every ten numbers you will see only two numbers divisible by three.
say in 10s-- 12,15
in 20s--21,24
in 30s--33,36
.
.
in 60s--63,66
so in every 100 numbers we will have 6*2 numbers following above rule.
in every 1000 numbers we will have 6*2*6 numbers following above rule.
since problem says four digits...6*2*6*6 numbers = 432 possibilities.

shankar.ashwin wrote:Different four-digit numbers are formed using the digits 1, 2, 3, 4, 5 and 6 (repetition is allowed). If 'n' numbers out of these are divisible by 3, then the value of 'n' is:

A) 234
B) 242
C) 324
D) 432
E) 464

One more approach.

The number of options for each digit = 6.
Thus, the total number of integers that can be formed = 6*6*6*6 = 1296.

Whatever digits are chosen for the 3 leftmost digits, the options for the units digit -- 1,2,3,4,5,6 -- will yield 6 consecutive integers.
For example, if the 3 leftmost digits are 111, the following integers can be formed:
1111,1112,1113,1114,1115,1116.
Since exactly 1 of every 3 consecutive integers is a multiple of 3, 1/3 of the total number of possible integers will be a multiple of 3.

(1/3)*1296 = 432.