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24 Feb 2020, 05:27
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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:
65%
(hard)
Question Stats:
58% (01:53) correct
42%
(01:45)
wrong
based on 674
sessions
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Date
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If n is a positive integer, what is the remainder when \(n^7 – n\) is divided by 42?
A. 0
B. 1
C. 3
D. 5
E. Cannot be determined.
Are You Up For the Challenge: 700 Level Questions
A. 0
B. 1
C. 3
D. 5
E. Cannot be determined.
Are You Up For the Challenge: 700 Level Questions
If n is a positive integer, what is the remainder when \(n^{7}–n\) is divided by 42?
42= 2*3*7
\(n^{7}–n= n*(n^6-1)= n*(n^3-1)(n^3+1)= n(n-1)(n^2+n+1)(n+1)(n^2-n+1)\\
\)
--> \(n(n-1)(n+1)(n^2+n+1)(n^2-n+1)= n(n-1)(n+1)(n^4+n^2+1)\)
Consecutive three integers \((n-1)(n)(n+1)\) can be divided by 6.
Also, we have to check whether 7 is a factor of \(n^7-n.\)
n -positive integer:
--> if \(n =1\), then n^7-n= 0
--> if \(n=2\), then \((n^4+n^2+1)= n^{2}(n^{2}+1) +1= 4*5+1= 21\) (yes)
--> if \(n=3\), then \(n^{2}(n^{2}+1) +1= 9*10+1= 91\) (yes)
--> if \(n=4\), then \(n^{2}(n^{2}+1) +1= 16*17 +1= 273= 7*39\) (yes )
......
In any positive integer of n, \(42\) is divided by \(n^{7}-n \)
The remainder will be \(0\)
The answer is A.
42= 2*3*7
\(n^{7}–n= n*(n^6-1)= n*(n^3-1)(n^3+1)= n(n-1)(n^2+n+1)(n+1)(n^2-n+1)\\
\)
--> \(n(n-1)(n+1)(n^2+n+1)(n^2-n+1)= n(n-1)(n+1)(n^4+n^2+1)\)
Consecutive three integers \((n-1)(n)(n+1)\) can be divided by 6.
Also, we have to check whether 7 is a factor of \(n^7-n.\)
n -positive integer:
--> if \(n =1\), then n^7-n= 0
--> if \(n=2\), then \((n^4+n^2+1)= n^{2}(n^{2}+1) +1= 4*5+1= 21\) (yes)
--> if \(n=3\), then \(n^{2}(n^{2}+1) +1= 9*10+1= 91\) (yes)
--> if \(n=4\), then \(n^{2}(n^{2}+1) +1= 16*17 +1= 273= 7*39\) (yes )
......
In any positive integer of n, \(42\) is divided by \(n^{7}-n \)
The remainder will be \(0\)
The answer is A.
General Discussion
25 Feb 2020, 08:29
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Bunuel
given \(n^7 – n\)
n(n^6-1)
n(n^6-1)/6*7 ; put n=0,1,2,3 we get remainder as 0
IMO A
