Is positive integer n-1 a multiple of 3?
(1) n³ - n is a multiple of 3
(2) n³ + 2n² + n is a multiple of 3
Statement 1:
n³ - n = n(n² - 1) = (n)(n+1)(n-1).
Case 1: n=2, with the result that (n)(n+1)(n-1) = 2*3*1.
In this case, n-1 = 2-1 = 1, which is NOT a multiple of 3.
Case 2: n=3, with the result that (n)(n+1)(n-1) = 3*4*2.
In this case, n-1 = 3-1 = 2, which is NOT a multiple of 3.
Case 3: n=4, with the result that (n)(n+1)(n-1) = 4*5*3.
In this case, n-1 = 4-1 = 3, which IS a multiple of 3.
Since n-1 is a multiple of 3 in Case 3 but not a multiple of 3 in Cases 1 and 2, INSUFFICIENT.
Statement 2:
n³ + 2n² + n = n(n² + 2n + 1) = (n)(n+1)(n+1).
Case 1: n=2, with the result that (n)(n+1)(n+1) = 2*3*3.
In this case, n-1 = 2-1 = 1, which is NOT a multiple of 3.
Case 2: n=3, with the result that (n)(n+1)(n+1) = 3*4*4.
In this case, n-1 = 3-1 = 2, which is NOT a multiple of 3.
n=4 implies that (n)(n+1)(n+1) = 4*5*5.
Since n=4 does not satisfy the constraint that (n)(n+1)(n+1) is a multiple of 3, n=4 is not a valid option in Statement 2.
Case 3: n=5, with the result that (n)(n+1)(n+1) = 5*6*6.
In this case, n-1 = 5-1 = 4, which is NOT a multiple of 3.
Case 4: n=6, with the result that (n)(n+1)(n+1) = 6*7*7.
In this case, n-1 = 6-1 = 5, which is NOT a multiple of 3.
In every case, n-1 is NOT a multiple of 3.
SUFFICIENT.
The correct answer is
B.
Last edited by
GMATGuruNY on Tue Aug 11, 2015 7:58 am, edited 2 times in total.
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