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.
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