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For every positive integer n, the highest number that n(n^2 – 1)(5n +

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For every positive integer n, the highest number that n(n^2 – 1)(5n +  [#permalink]

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New post 19 Feb 2020, 01:23
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Question Stats:

48% (01:51) correct 53% (02:08) wrong based on 39 sessions

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Re: For every positive integer n, the highest number that n(n^2 – 1)(5n +  [#permalink]

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New post 19 Feb 2020, 02:12
2
(n–1)*n*(n+1)(5n + 2)
n=2 --> 1*2*3*12 divisible by 6, 24 and 36
n=3 --> 2*3*4*17 divisible by 6, 24, but NOT 36

FINAL ANSWER IS (B) 24
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Re: For every positive integer n, the highest number that n(n^2 – 1)(5n +  [#permalink]

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New post 19 Feb 2020, 03:30
1
Quote:
For every positive integer n, the highest number that n(n^2 – 1)(5n + 2) is always divisible by is

A. 6
B. 24
C. 36
D. 48
E. 96


The smartest way in such problems is to substitute values and observe the results. Always try with three consecutive values

@n=1, n(n^2 – 1)(5n + 2) = 0
@n=2, n(n^2 – 1)(5n + 2) = 2*3*12 = 72
@n=3, n(n^2 – 1)(5n + 2) = 3*8*17 = 24*17
@n=4, n(n^2 – 1)(5n + 2) = 4*15*22 = 24*5*11
@n=5, n(n^2 – 1)(5n + 2) = 5*24*27

The common and highest number (GCD) that divides all the results obtained = 24

Answer: Option B
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Re: For every positive integer n, the highest number that n(n^2 – 1)(5n +  [#permalink]

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New post 19 Feb 2020, 04:29
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Solution:
n (n^2 – 1) (5n + 2)
= (n – 1), n,(n + 1), (5n +2)
Putting n = 2, 3, 4, 5,………….
n =2,
(n – 1)* n*(n + 1)*(5n +2)
= 1* 2* 3* 12, divisible by 24

n = 3,
(n – 1)* n*(n + 1)*(5n +2)
=2* 3* 4 *17, divisible by 24

n = 4,
(n – 1)* n*(n + 1)*(5n +2)
= 3*4*5*22, divisible by 24

n =5,
(n – 1)* n*(n + 1)*(5n +2)
= 4*5*6*27, divisible by 24
Answer : B
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Re: For every positive integer n, the highest number that n(n^2 – 1)(5n +  [#permalink]

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New post 19 Feb 2020, 07:12
1
Quote:
For every positive integer n, the highest number that n(n^2 – 1)(5n + 2) is always divisible by is

A. 6
B. 24
C. 36
D. 48
E. 96


n is any positive integer
n(n^2 – 1)(5n + 2)
n(n+1)(n-1)(5n + 2)
for n=odd=1: 1(1+1)(1-1)(5+2)=0
for n=odd=3: 3(4)(2)(17)
for n=even=2: 2(3)(1)(12)
gcf(for n: 3,2): 3*2*4=24

note: product of three consecutive integers is divisible by at least two evens, and one odd.

Ans (B)
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Re: For every positive integer n, the highest number that n(n^2 – 1)(5n +  [#permalink]

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New post 19 Feb 2020, 21:31
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X:n(n^2 – 1)(5n + 2)

starting with
n=1....X=0
n=2....X=2(3)(12)=24*3
n=3....X=3(8)(17)=24*17
n=4....X=4(15)(22)= 24*55

the greatest common divisor is 24

OA:B
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Re: For every positive integer n, the highest number that n(n^2 – 1)(5n +  [#permalink]

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New post 19 Feb 2020, 03:56
Substitute with the smallest positive integer where the answer is not 0
when we use 2 we get the answer 72 and highest divisible integer is 36

(c)
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Re: For every positive integer n, the highest number that n(n^2 – 1)(5n +  [#permalink]

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New post 19 Feb 2020, 04:34
For every positive integer n, the highest number that n(n^2 – 1)(5n + 2) is always divisible by is

A. 6
B. 24
C. 36
D. 48
E. 96

(n-1)*n*(n+1) -> 3 consecutive numbers -> will always be divisible by 6

5n+2 can be 7,12,17 etc

So the equation will always be divisible by 6 as the highest number

Answer - A
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Re: For every positive integer n, the highest number that n(n^2 – 1)(5n +  [#permalink]

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New post 19 Feb 2020, 08:25
n(n^2 – 1)(5n + 2) ; n( n+1)(n-1)(5n+2)

for n =1 ; we get 0
n=2; 2*3*1*12 ; 2^3*3^2
n=3 ; 2^3*3*17
common is for all values 2^3*3 ; 24
IMO B ; 24



For every positive integer n, the highest number that n(n^2 – 1)(5n + 2) is always divisible by is

A. 6
B. 24
C. 36
D. 48
E. 96
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Re: For every positive integer n, the highest number that n(n^2 – 1)(5n +  [#permalink]

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New post 19 Feb 2020, 10:18
For every positive integer n, the highest number that n(n^2 – 1)(5n + 2) is always divisible by is

A. 6
B. 24
C. 36
D. 48
E. 96

\(n(n^2 – 1)(5n + 2)\) = (n – 1)n(n + 1)(5n + 2)
which suggests that it is always divisible by 3 and 2 since (n – 1)n(n + 1) is a multiple of three consecutive numbers.

Hence 6 is the highest number that divides \(n(n^2 – 1)(5n + 2)\) always
Also, if least values of n are considered, the highest number can be found out.

For 1*2*3... , 2*3*4.. , 3*4*5... 6 is the number that divides the \(n(n^2 – 1)(5n + 2)\) always.

Answer A.
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Re: For every positive integer n, the highest number that n(n^2 – 1)(5n +   [#permalink] 19 Feb 2020, 10:18

For every positive integer n, the highest number that n(n^2 – 1)(5n +

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