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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.

Re: Is x divisible by 30? [#permalink]
31 Jan 2012, 16:23

14

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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.

Re: Is x divisible by 30? [#permalink]
01 Feb 2012, 13:38

Bunuel wrote:

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.

Hi Bunnel,

I've a doubt in the following:

(n-1)n(n+1) ..is the product of any 3 consecutive integers divisble by 3 or 6 (3!)? (I came across the following rule in this forum : product of n consecutive integers is divisble by n!.)

If the product is divisible by 6 , then we need to check for only divisbility by 5?

Re: Is x divisible by 30? [#permalink]
01 Feb 2012, 13:44

Expert's post

anuu wrote:

Hi Bunnel,

I've a doubt in the following:

(n-1)n(n+1) ..is the product of any 3 consecutive integers divisble by 3 or 6 (3!)? (I came across the following rule in this forum : product of n consecutive integers is divisble by n!.)

If the product is divisible by 6 , then we need to check for only divisbility by 5?

Regards, Anu

In my solution I showed that n^5-n is divisible by 10 and 3, so by 30. You can do in another way too: prove divisibility by 5 and divisibility by 6. _________________

Re: Is x divisible by 30? [#permalink]
27 May 2013, 10:56

2

This post received KUDOS

Expert's post

enigma123 wrote:

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

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. _________________

Re: Is x divisible by 30? [#permalink]
30 Oct 2013, 08:30

Bunuel wrote:

anuu wrote:

Hi Bunnel,

I've a doubt in the following:

(n-1)n(n+1) ..is the product of any 3 consecutive integers divisble by 3 or 6 (3!)? (I came across the following rule in this forum : product of n consecutive integers is divisble by n!.)

If the product is divisible by 6 , then we need to check for only divisbility by 5?

Regards, Anu

In my solution I showed that n^5-n is divisible by 10 and 3, so by 30. You can do in another way too: prove divisibility by 5 and divisibility by 6.

Hi How do you prove divisibility by 5 and 6? Is it possible?

Re: Is x divisible by 30? [#permalink]
30 Oct 2013, 08:37

Expert's post

ronr34 wrote:

Bunuel wrote:

anuu wrote:

Hi Bunnel,

I've a doubt in the following:

(n-1)n(n+1) ..is the product of any 3 consecutive integers divisble by 3 or 6 (3!)? (I came across the following rule in this forum : product of n consecutive integers is divisble by n!.)

If the product is divisible by 6 , then we need to check for only divisbility by 5?

Regards, Anu

In my solution I showed that n^5-n is divisible by 10 and 3, so by 30. You can do in another way too: prove divisibility by 5 and divisibility by 6.

Hi How do you prove divisibility by 5 and 6? Is it possible?

Please read the solution carefully:

(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. _________________

Re: Is x divisible by 30? [#permalink]
16 Dec 2014, 10:54

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