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If the three unique positive digits A,B, and C are arranged in all pos
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09 Jul 2021, 08:30
09 Jul 2021, 08:30
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If the three unique positive digits A,B, and C are arranged in all possible sequences (ABC,ACB,BAC,ETC....),then the sum of all the resulting three -digit integers must be divisible by each of the following EXCEPT?
(A) 2
(B) 3
(C) 6
(D) 11
(E) 37
(A) 2
(B) 3
(C) 6
(D) 11
(E) 37
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Re: If the three unique positive digits A,B, and C are arranged in all pos
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10 Jul 2021, 22:36
10 Jul 2021, 22:36
2
GeminiHeat wrote:
If the three unique positive digits A,B, and C are arranged in all possible sequences (ABC,ACB,BAC,ETC....),then the sum of all the resulting three -digit integers must be divisible by each of the following EXCEPT?
(A) 2
(B) 3
(C) 6
(D) 11
(E) 37
(A) 2
(B) 3
(C) 6
(D) 11
(E) 37
Remember: Any 3 digit number \(xyz\) can be written as \(100x + 10y + z\)
Now, ABC can be arranged in 3! ways = 6 ways
i.e. ABC, ACB, BAC, BCA, CAB, CBA
ABC = 100A + 10B + C
ACB = 100A + 10C + B
BAC = 100B + 10A + C
BCA = 100B + 10C + A
CAB = 100C + 10A + B
CBA = 100C + 10B + A
Sum = 222A + 222B + 222C = 222(A + B + C) = (2)(3)(37)(A + B + C)
Possible factors = 1, 2, 3, 6, 37, 74, 111, 222
Hence, option D
Re: If the three unique positive digits A,B, and C are arranged in all pos
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07 Mar 2024, 04:19
07 Mar 2024, 04:19
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