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02 Feb 2022, 06:27
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
56% (02:04) correct
44%
(02:04)
wrong
based on 126
sessions
History
Date
Time
Result
Not Attempted Yet
How many six-digit positive integers comprising only the digits 1 or 2 can be formed such that the number is divisible by 3?
A. 3
B. 20
C. 22
D. 38
E. 360
A. 3
B. 20
C. 22
D. 38
E. 360
03 Feb 2022, 03:22
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Divisibility rule for 3: If sum of digits is divisible by 3, then the the number is divisible by 3.
There are only three ways in which this works:
1 - all the digits are 1
2- all the digits are 2
3 - three digits are 1 and three digits are 2
1: 111111 can only be arranged in that single way.
2: 222222 can only be arranged in that single way.
3: 111222 can be arranged in \(\frac{6!}{3!3!} = 20\) ways
Total: \(1 + 1 + 20 = 22\)
Answer C
There are only three ways in which this works:
1 - all the digits are 1
2- all the digits are 2
3 - three digits are 1 and three digits are 2
1: 111111 can only be arranged in that single way.
2: 222222 can only be arranged in that single way.
3: 111222 can be arranged in \(\frac{6!}{3!3!} = 20\) ways
Total: \(1 + 1 + 20 = 22\)
Answer C
09 Jan 2025, 02:15
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