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M08-35

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Bunuel
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M08-35 [#permalink] New post   16 Sep 2014, 00:38
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82% (02:01) correct 18% (02:14) wrong based on 89 sessions
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For all positive integers \(m\), \(\&m = m - 1\) when \(m\) is odd and \(\&m = m + 2\) when \(m\) is even. What is the value of \(\&(...\&(\&(\&(15)))...)\), where \(\&\) is used 99 times?


A. 120
B. 180
C. 210
D. 225
E. 250
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C
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Bunuel
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Re M08-35 [#permalink] New post   16 Sep 2014, 00:38
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For all positive integers \(m\), \(\&m = m - 1\) when \(m\) is odd and \(\&m = m + 2\) when \(m\) is even. What is the value of \(\&(...\&(\&(\&(15)))...)\), where \(\&\) is used 99 times?


A. 120
B. 180
C. 210
D. 225
E. 250


Since 15 is odd, then \(\&15 = 15-1=14\).

Now, we need to calculate \(\&(...\&(\&(\&(14)))...)\), where \(\&\) is used 98 times.

Since 14 is even, each \(\&\) just adds 2 to the previous result. Therefore, \(\&(...\&(\&(\&(14)))...)\), where \(\&\) is used 98 times, equals to \(14 + 2*98 = 210\).


Answer: C
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Re: M08-35 [#permalink] New post   27 Jun 2023, 01:37
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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M08-35 [#permalink] New post   22 Sep 2023, 06:25
Bunuel
Since we have an AP series (14,16..) after the first operation. On using the formula for nth term. I am getting the nth term as 208. (where the first term is 14, d=2 and n=98.). What is it that I am missing out?
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Re: M08-35 [#permalink] New post   22 Sep 2023, 08:19
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ifyouknowyouknow wrote:
Bunuel
Since we have an AP series (14,16..) after the first operation. On using the formula for nth term. I am getting the nth term as 208. (where the first term is 14, d=2 and n=98.). What is it that I am missing out?


Including 14, n comes out to be 99 not 98. You can try smaller number of & to test your logic.
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Re: M08-35 [#permalink]
22 Sep 2023, 08:19
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Bunuel
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