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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:
45% (02:51) correct
55%
(02:29)
wrong
based on 175
sessions
History
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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
07 Feb 2014, 05:09
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Is x divisible by 30?
(1) x = k*(m^3 - m), where m and k are both integers > 9 --> x=k*(m-1)m(m+1). Well, if k=30 then the answer is YES, but if k=11, then (m-1)m(m+1) only guarantees divisibility by 6, so the answer might be NO (for example m=12). Not sufficient.
(2) x = n^5 - n, where n is an integer > 9 --> the last digit of any integer in positive integer power repeats in pattern of 4 (some numbers have the pattern of 2 or 1, but pattern of 4 holds for them too). So, n^5 has the same last digit as n^1 or simply n, therefore the last digit of n^5-n is ALWAYS zero, so its ALWAYS divisible by 10. Next, n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=(n-1)n(n+1)(n^2+1) --> (n-1)n(n+1) is divisible by 3 (as there are 3 consecutive integers) --> n^5-n is divisible by 10*3=30. Sufficient.
Answer: B.
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 --> x=k*(m-1)m(m+1). Well, if k=30 then the answer is YES, but if k=11, then (m-1)m(m+1) only guarantees divisibility by 6, so the answer might be NO (for example m=12). Not sufficient.
(2) x = n^5 - n, where n is an integer > 9 --> the last digit of any integer in positive integer power repeats in pattern of 4 (some numbers have the pattern of 2 or 1, but pattern of 4 holds for them too). So, n^5 has the same last digit as n^1 or simply n, therefore the last digit of n^5-n is ALWAYS zero, so its ALWAYS divisible by 10. Next, n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=(n-1)n(n+1)(n^2+1) --> (n-1)n(n+1) is divisible by 3 (as there are 3 consecutive integers) --> n^5-n is divisible by 10*3=30. Sufficient.
Answer: B.
OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-divisible-by-126857.html
General Discussion
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.
