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There is a sequence An for a positive integer n such that when An-2 is

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MathRevolution
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There is a sequence An for a positive integer n such that when An-2 is [#permalink] New post   19 Apr 2016, 06:02
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There is a sequence An for a positive integer n such that when An-2 is divided by An-1 the remainder is An. If A3=6, A4=0, which of the following can be the value of A1?
Attachment:
positive integer.jpg
positive integer.jpg [ 14.57 KiB | Viewed 4396 times ]

A. 48
B. 50
C. 52
D. 56
E. 58


* A solution will be posted in two days.
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There is a sequence An for a positive integer n such that when An-2 is [#permalink] New post   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
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Re: There is a sequence An for a positive integer n such that when An-2 is [#permalink] New post   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
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Re: There is a sequence An for a positive integer n such that when An-2 is [#permalink] New post   21 Apr 2016, 01:26
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There is a sequence An for a positive integer n such that when An-2 is divided by An-1 the remainder is An. If A3=6, A4=0, which of the following can be the value of A1?

A. 48
B. 50
C. 52
D. 56
E. 58


==> If n=4, A2=A3Q+A4=6Q+0=6Q is derived.
If n=3, from A1=A2P+A3=6QP+6=6(QP+1), it becomes always a multiple of 6.
Thus, the answer is A.
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Re: There is a sequence An for a positive integer n such that when An-2 is [#permalink] New post   13 Oct 2019, 10:40
MathRevolution wrote:
There is a sequence An for a positive integer n such that when An-2 is divided by An-1 the remainder is An. If A3=6, A4=0, which of the following can be the value of A1?
Attachment:
positive integer.jpg

A. 48
B. 50
C. 52
D. 56
E. 58


* A solution will be posted in two days.

Can you please clarify this?
Does the question imply that when A1/A2 the remainder is A3. If yes then A3=6, which means when A1/A2 the remainder should be 6. --> Since, when A2/A3 , remainder is A4=0, A2should be also a multiple of 6. --> when A1 / A2( multiple of 6) remainder should be 6 as per q. If 48 / multiple of 6, remainder is 0 and not 6. I am confused.
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There is a sequence An for a positive integer n such that when An-2 is [#permalink] New post   19 Oct 2019, 21:24
MathRevolution wrote:
There is a sequence An for a positive integer n such that when An-2 is divided by An-1 the remainder is An. If A3=6, A4=0, which of the following can be the value of A1?
Attachment:
positive integer.jpg

A. 48
B. 50
C. 52
D. 56
E. 58


* A solution will be posted in two days.


48 can't be the answer.

Here A4=A2/A3
Hence A2 should be a multiple of 6.(6,12,18,24....so on).

Now A3=A1/A2

Here A3=6(reminder) and A2=6 or 12 or 18 and so on

Therefore reminder should be 6 when A1 is divided by 6k.

In none on the mentioned options above condition gets satisfied.

Kindly correct where I am wrong.
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Re: There is a sequence An for a positive integer n such that when An-2 is [#permalink] New post   22 Mar 2024, 10:37
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