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15 Feb 2024, 09:33
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Dropdown 1: 4
Dropdown 2: 42
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:
63% (02:50) correct
37%
(02:49)
wrong
based on 1358
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Data Insights (DI) Butler 2023-24 [Question #222, Date: Jan-30-2024] [Click here for Details]
If N = 5 is the starting value, then after the process "N =N/2" is completed for the second time, the output is
Like N= 10,N = is a number that passes only once through the process "N = 3N + 1" before it reaches "stop."
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Official Answer
Dropdown 1: 4
Dropdown 2: 42
16 Feb 2024, 01:35
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IMO 4 & 42
Logic-
For Statement 1,
If N = 5 is the starting value, then after the process "N =N/2" is completed for the second time, the output is
Is N = 1? No ---> Is N Odd? Yes ---> N=3N+1 i.e. N=3*5+1=16--->N=N/2 i.e. N=16/2=8. 1st Cycle is complete.
With N=8, Is N = 1? No ---> Is N Odd? No ---> N=N/2 i.e. N=8/2=4. 2nd Cycle is complete and 4 should be our answer.
For Statement 2,
With Trial and error we can find that N=42 is a number that passes only once through the process "N = 3N + 1" before it reaches "stop."
Logic-
For Statement 1,
If N = 5 is the starting value, then after the process "N =N/2" is completed for the second time, the output is
Is N = 1? No ---> Is N Odd? Yes ---> N=3N+1 i.e. N=3*5+1=16--->N=N/2 i.e. N=16/2=8. 1st Cycle is complete.
With N=8, Is N = 1? No ---> Is N Odd? No ---> N=N/2 i.e. N=8/2=4. 2nd Cycle is complete and 4 should be our answer.
For Statement 2,
With Trial and error we can find that N=42 is a number that passes only once through the process "N = 3N + 1" before it reaches "stop."
30 Mar 2024, 22:07
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1. If N = 5 is the starting value, then after the process "N =N/2" is completed for the second time, the output is [4]
n=5
=> \(n\neq{1}\)
=> n ood
=> n = 3 * 5 + 1 = 16
=> n = \(\frac{16}{2}\) = 8 (1st time)
=> \(n\neq{1}\)
=> n not odd
=> n = \(\frac{8}{2}\) = 4 (2nd time)
Answer: 4
2. Like N= 10,N =[42] is a number that passes only once through the process "N = 3N + 1" before it reaches "stop."
n = 10
=> \(n\neq{1}\)
=> n not odd
=> n = 5
... (like above)
=> n = 8 (n = \(\frac{n}{2}\) for the 1st time)
...
=> n = 4 (n = n/2 for the 1st time)
=> \(n\neq{1}\)
=> n not odd
=> n = 2
... (one more time divided by 2)
=> n = 1
=> The patterns seems that after the 1st time going thru "n=3n+1", the results must be something \(2^x\) to keep dividing by 2 without going thru "n=3n+1" again
Likely to choose an even to start trying (because the first N must go thru the "odd-even" test first)
42
=> n = \(\frac{42}{2}\) = 21
=> n = 21*3+1 = 64 = 2^6
Answer: 42
n=5
=> \(n\neq{1}\)
=> n ood
=> n = 3 * 5 + 1 = 16
=> n = \(\frac{16}{2}\) = 8 (1st time)
=> \(n\neq{1}\)
=> n not odd
=> n = \(\frac{8}{2}\) = 4 (2nd time)
Answer: 4
2. Like N= 10,N =[42] is a number that passes only once through the process "N = 3N + 1" before it reaches "stop."
n = 10
=> \(n\neq{1}\)
=> n not odd
=> n = 5
... (like above)
=> n = 8 (n = \(\frac{n}{2}\) for the 1st time)
...
=> n = 4 (n = n/2 for the 1st time)
=> \(n\neq{1}\)
=> n not odd
=> n = 2
... (one more time divided by 2)
=> n = 1
=> The patterns seems that after the 1st time going thru "n=3n+1", the results must be something \(2^x\) to keep dividing by 2 without going thru "n=3n+1" again
Likely to choose an even to start trying (because the first N must go thru the "odd-even" test first)
42
=> n = \(\frac{42}{2}\) = 21
=> n = 21*3+1 = 64 = 2^6
Answer: 42
