Source: Manhattan GMAT
Is positive integer n – 1 a multiple of 3?
(1) n^3 – n is a multiple of 3
(2) n^3 + 2n^2+ n is a multiple of 3
OA in a few, please explain. THanks:)
n-1 a multiple of 3?
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Hey, maxim - can you check your formatting please? Please indicate any exponents with a carat (eg x^3). Thanks!
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Hi maxim 730,
let's try to guess if positive integer n – 1 a multiple of 3?
REGRADING THE QUESTIONS FIRST ASSUMPTION :
(1) n3 – n is a multiple of 3 : this means that n(n-1)(n+1) IS MULTIPLE OF 3 : this means that each one could suffisantly be a multiple of three. So the assumption 1 is insuffisant ( so answer is either B, D or E).
To make sure of this first answer, you can pick n = 2. or n=3; both possibilities give you a number of ( n^3 - n) that is multiple of 3;
REGARDING THE ASSUMPTION 2: (2) says that n3 + 2n2+ n is a multiple of 3
this leads to n(n+1)^2 is multiple of 3 , we, here, have two possibilities, either n is multiple of 3 or (n+1)^2 is multiple of 3
let's take look at each of these conclusions :
* if n is multiple of 3, then n-1 is not multiple of 3 ( to rephrase, we would say that the remainder of the division of n-1 by 3 is 2)
* If (n+1)^2 is multiple of 3, ( to rephrase, 3 is prime factor of (n+1)2) , then necessarely, n+1 is multilpe 3, because if it wasn't not a prime factor of (n+1), it would'nt be a factor of (n+1)^2, ALL THAT means that (n+1) is multiple of 3, and consequently ( n-1) is not a multiple of 3.
So I would go for B;
I appplogize for my wordy and awkward sentences, you would understand that english my third langage.
PLEASE CAN ANYBODY COMMENT MY REASONING,
good luck for us.
let's try to guess if positive integer n – 1 a multiple of 3?
REGRADING THE QUESTIONS FIRST ASSUMPTION :
(1) n3 – n is a multiple of 3 : this means that n(n-1)(n+1) IS MULTIPLE OF 3 : this means that each one could suffisantly be a multiple of three. So the assumption 1 is insuffisant ( so answer is either B, D or E).
To make sure of this first answer, you can pick n = 2. or n=3; both possibilities give you a number of ( n^3 - n) that is multiple of 3;
REGARDING THE ASSUMPTION 2: (2) says that n3 + 2n2+ n is a multiple of 3
this leads to n(n+1)^2 is multiple of 3 , we, here, have two possibilities, either n is multiple of 3 or (n+1)^2 is multiple of 3
let's take look at each of these conclusions :
* if n is multiple of 3, then n-1 is not multiple of 3 ( to rephrase, we would say that the remainder of the division of n-1 by 3 is 2)
* If (n+1)^2 is multiple of 3, ( to rephrase, 3 is prime factor of (n+1)2) , then necessarely, n+1 is multilpe 3, because if it wasn't not a prime factor of (n+1), it would'nt be a factor of (n+1)^2, ALL THAT means that (n+1) is multiple of 3, and consequently ( n-1) is not a multiple of 3.
So I would go for B;
I appplogize for my wordy and awkward sentences, you would understand that english my third langage.
PLEASE CAN ANYBODY COMMENT MY REASONING,
good luck for us.
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(1) n^3 - n is a multiple of 3
(2) n^3 + 2n^2+ n is a multiple of 3
Option 1...
n^3 -n = n(n^2 - 1) = n (N^2 - 1^2) = n (n + 1) ( n - 1)
Option 2 ...
n^3 + 2n^2+ n = n(n^2 + 2n + 1) = n (n+1) ( n + 1)
now from option (2) . either n or (n+1) is multiple of 3.
in any case n - 1 would not be a multiple of 3.
(2) n^3 + 2n^2+ n is a multiple of 3
Option 1...
n^3 -n = n(n^2 - 1) = n (N^2 - 1^2) = n (n + 1) ( n - 1)
Option 2 ...
n^3 + 2n^2+ n = n(n^2 + 2n + 1) = n (n+1) ( n + 1)
now from option (2) . either n or (n+1) is multiple of 3.
in any case n - 1 would not be a multiple of 3.
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