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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
(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
As OA is not given, I tried it as below and for me the answer should be B. Any help in correcting or confirming the answer will be very much appreciated.
Question Says
is x divisible by 30 i.e. does x have atleast 2,3 and 5 as its Prime factors?
Considering Statement 1
x= k [m(m^2-1)] ------------------ (After factorization)------------------(1)
x = k (m-1,m and m+1) --------------------------------------------------(2)
From 2 we can tell that m-1,m and m+1 are consecutive integers. But say m = 17 then (m-1) (m) (m+1) is not divisible by 30 and also we don't know the value of K. But when m = 10 then (m-1) (m) (m+1) is divisible by 30 and irrespective of the value of K it will always be divisible by 30. Therefore this statement alone is insufficient to answer this question.
Considering statement 2
x = \(n^5\)-n
By factorization it can be simplified to \(n^2\) (n+1) (n-1). No matter whether n will be ODD or EVEN this will always be divisible by 30 because n^2 (n+1) (n-1) will always have more prime factors than 30. Therefore sufficient and hence B is my answer.
Question Says
is x divisible by 30 i.e. does x have atleast 2,3 and 5 as its Prime factors?
Considering Statement 1
x= k [m(m^2-1)] ------------------ (After factorization)------------------(1)
x = k (m-1,m and m+1) --------------------------------------------------(2)
From 2 we can tell that m-1,m and m+1 are consecutive integers. But say m = 17 then (m-1) (m) (m+1) is not divisible by 30 and also we don't know the value of K. But when m = 10 then (m-1) (m) (m+1) is divisible by 30 and irrespective of the value of K it will always be divisible by 30. Therefore this statement alone is insufficient to answer this question.
Considering statement 2
x = \(n^5\)-n
By factorization it can be simplified to \(n^2\) (n+1) (n-1). No matter whether n will be ODD or EVEN this will always be divisible by 30 because n^2 (n+1) (n-1) will always have more prime factors than 30. Therefore sufficient and hence B is my answer.
31 Jan 2012, 17:23
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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.
P.S. You have factored n^5-n incorrectly and also your reasoning for n^2(n+1)(n-1) is not right: it's not always divisible by 30, try n=12 for example.
Hope it's clear.
(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.
P.S. You have factored n^5-n incorrectly and also your reasoning for n^2(n+1)(n-1) is not right: it's not always divisible by 30, try n=12 for example.
Hope it's clear.
27 May 2013, 11:56
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enigma123
Bunuel has already explained the second part. Just another approach :
From F.S 1, we know that \(x = k*(m-1)*m*(m+1)\). Now, product of any 3 consecutive integers is ALWAYS divisible by 3.Also, the product of any two consecutive integers is ALWAYS divisible by 2. Thus, x is divisibe by 2 and 3--> x is divisible by atleast 6. Now we have to find out whether x is divisible by 5 or not.For k=m=10, it is. For k=m=13, it isn't.Insufficient.
From F.S 2, we know that x =\(n^5-n\) = \(n(n^2-1)(n^2+1)\) = \((n-1)(n)(n+1)(n^2+1)\). Just as above, we know that x is atleast divisibly by 6.Again, we have to find out whether it is divisibly by 5 or not-->If there was some way in which I could represent \((n^5-n)\) as a product of 5 consecutive integers, then I would be succesfull in proving that \(n^5-n\) is divisibly by 5 also.
Now, assuming n as the median,to get a product of 5 consecutive integers, we would need (n-2) and (n+2)-->\((n^2-4)\)-->\((n-1)n(n+1)[(n^2-4)+(4+1)]\) = (n-2)(n-1)n(n+1)(n+2) + 5(n-1)n(n+1)--> This will be always be divisible by 30.
There is actually a very famous theorem which states that \(x^p-x\) is always divisible by p, provided p is prime and x an integer. However, it is not relevant w.r.t GMAT.
