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13 Mar 2008, 08:57
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
46% (02:51) correct
54%
(02:30)
wrong
based on 182
sessions
History
Date
Time
Result
Not Attempted Yet
Is x divisible by 30?
(1) x = k(m^3 - m), where m and k are both integers > 9
(2) x = n^5 - n, where n is an integer > 9
OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-divisible-by-126857.html
(1) x = k(m^3 - m), where m and k are both integers > 9
(2) x = n^5 - n, where n is an integer > 9
OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-divisible-by-126857.html
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
13 Mar 2008, 10:11
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B
\(30=2*3*5\)
1. \(x = k(m^3 - m)=k*m*(m^2-1)=k*m*(m-1)(m+1)\)
\((m-1),m,(m+1)\) are consecutive integers and the product of these integers will always be divisible by 6 (2 and 3).
But the consequence can be divisible by 5 or cannot be. Therefore, if k and the product are not divisible by 5, x will not be divisible by 30; if k or the product is divisible by 5, x will be divisible by 30. INSUFF.
2. \(x = n^5 - n=n*(n^4-1)=n*(n^2-1)*(n^2+1)=n*(n-1)*(n+1)*(n^2+1)\)
\((n-1),n,(n+1)\) are consecutive integers and the product of these integers will always be divisible by 6 (2 and 3).
if n=5k, 5k+1, or 5k+4, the product will be divisible by 5 and therefore, by 30.
if n=5k+2 or 5k+3 the product will not be divisible by 5.
but: n=5k+2: \((n^2+1)=(5k+2)^2+1=25k^2+20k+5=5*(5k^2+4k+1)\) Therefore, x is divisible by 30
n=5k+3: \((n^2+1)=(5k+3)^2+1=25k^2+30k+10=5*(5k^2+6k+2)\) Therefore, x is also divisible by 30.
SUFF.
\(30=2*3*5\)
1. \(x = k(m^3 - m)=k*m*(m^2-1)=k*m*(m-1)(m+1)\)
\((m-1),m,(m+1)\) are consecutive integers and the product of these integers will always be divisible by 6 (2 and 3).
But the consequence can be divisible by 5 or cannot be. Therefore, if k and the product are not divisible by 5, x will not be divisible by 30; if k or the product is divisible by 5, x will be divisible by 30. INSUFF.
2. \(x = n^5 - n=n*(n^4-1)=n*(n^2-1)*(n^2+1)=n*(n-1)*(n+1)*(n^2+1)\)
\((n-1),n,(n+1)\) are consecutive integers and the product of these integers will always be divisible by 6 (2 and 3).
if n=5k, 5k+1, or 5k+4, the product will be divisible by 5 and therefore, by 30.
if n=5k+2 or 5k+3 the product will not be divisible by 5.
but: n=5k+2: \((n^2+1)=(5k+2)^2+1=25k^2+20k+5=5*(5k^2+4k+1)\) Therefore, x is divisible by 30
n=5k+3: \((n^2+1)=(5k+3)^2+1=25k^2+30k+10=5*(5k^2+6k+2)\) Therefore, x is also divisible by 30.
SUFF.
13 Mar 2008, 10:13
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terp26
1: x = k(m^3 - m)
x = k m (m^2 - 1)
x = k m (m - 1) (m+1)
x = (k) (m - 1) (m) (m+1)
so x is a product of k and 3 consequtive integers (m-1) (m) and (m+1).
if k = 11 and m = 13, x is not divisible by 30.
if k = 10 and m = 13, x is divisible by 30. so nsf.
2: x = (n^5 - n) and integer n is >9,
x = n (n-1) (n+1) (n^2 + 1)
in this case, no matter the value of n, x is divisible by 30.
if n is, lets say, 12, x = 11(12) (13) (145) suff.
if n is, lets say, 13, x = 13 (12) (14) (13^2 + 1) = 12x13x14x140. suff.
if n is, lets say, 17, x = 16(17) (18) (290). so suff.
so B.
