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19 Apr 2016, 06:02
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
72% (02:52) correct
28%
(03:03)
wrong
based on 590
sessions
History
Date
Time
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There is a sequence \(a_n\) for a positive integer n such that when \(a_{n-2}\) is divided by \(a_{n-1}\) the remainder is \(a_n\). If \(a_3=6\), \(a_4=0\), which of the following can be the value of \(a_1\)?
A. 48
B. 50
C. 52
D. 56
E. 58
positive integer.jpg [ 14.57 KiB | Viewed 10051 times ]
A. 48
B. 50
C. 52
D. 56
E. 58
Attachment:
positive integer.jpg [ 14.57 KiB | Viewed 10051 times ]
19 Apr 2016, 07:54
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When A1/A2, Remainder = A3 = 6 --> A1 = xA2 + 6
When A2/A3, Remainder = A4 = 0 --> A2 = yA3
A1 = xyA3 + 6 = kA3 + 6 = 6(k + 1) --> A1 must be a multiple of 6
Only option A is a multiple of 6
Answer: A
When A2/A3, Remainder = A4 = 0 --> A2 = yA3
A1 = xyA3 + 6 = kA3 + 6 = 6(k + 1) --> A1 must be a multiple of 6
Only option A is a multiple of 6
Answer: A
19 Apr 2016, 15:43
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An-2 =(An-1)Q + An
A2 = (A3)Q + A4
A2 = 6Q + 0 = 6Q
therefore
A1 = (6Q)Q + 6 = 6Q^2 + 6 = 6(Q^2 + 1)
Thus A1 is definitely divisible by 6
Out of the given options, only A is divisible by 6
Correct Option : A
A2 = (A3)Q + A4
A2 = 6Q + 0 = 6Q
therefore
A1 = (6Q)Q + 6 = 6Q^2 + 6 = 6(Q^2 + 1)
Thus A1 is definitely divisible by 6
Out of the given options, only A is divisible by 6
Correct Option : A
