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Is x divisible by 30? [#permalink]
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Updated on: 12 Jul 2013, 10:46
Question Stats:
44% (01:30) correct 56% (01:44) wrong based on 331 sessions
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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 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^21)]  (After factorization)(1) x = k (m1,m and m+1) (2)
From 2 we can tell that m1,m and m+1 are consecutive integers. But say m = 17 then (m1) (m) (m+1) is not divisible by 30 and also we don't know the value of K. But when m = 10 then (m1) (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) (n1). No matter whether n will be ODD or EVEN this will always be divisible by 30 because n^2 (n+1) (n1) will always have more prime factors than 30. Therefore sufficient and hence B is my answer.
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MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730
Originally posted by enigma123 on 31 Jan 2012, 16:52.
Last edited by Bunuel on 12 Jul 2013, 10:46, edited 1 time in total.
Added the OA.



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Re: Is x divisible by 30? [#permalink]
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31 Jan 2012, 17:23
Is x divisible by 30?(1) x = k*(m^3  m), where m and k are both integers > 9 > x=k*(m1)m(m+1). Well, if k=30 then the answer is YES, but if k=11, then (m1)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^5n is ALWAYS zero, so its ALWAYS divisible by 10. Next, n^5n=n(n^41)=n(n^21)(n^2+1)=(n1)n(n+1)(n^2+1) > (n1)n(n+1) is divisible by 3 (as there are 3 consecutive integers) > n^5n is divisible by 10*3=30. Sufficient. Answer: B. P.S. You have factored n^5n incorrectly and also your reasoning for n^2(n+1)(n1) is not right: it's not always divisible by 30, try n=12 for example. Hope it's clear.
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Re: Is x divisible by 30? [#permalink]
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31 Jan 2012, 17:26
Too good, Superb explanation and thanks for correcting me Bunuel.
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Re: Is x divisible by 30? [#permalink]
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01 Feb 2012, 14:38
Bunuel wrote: Is x divisible by 30?
(1) x = k*(m^3  m), where m and k are both integers > 9 > x=k*(m1)m(m+1). Well, if k=30 then the answer is YES, but if k=11, then (m1)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^5n is ALWAYS zero, so its ALWAYS divisible by 10. Next, n^5n=n(n^41)=n(n^21)(n^2+1)=(n1)n(n+1)(n^2+1) > (n1)n(n+1) is divisible by 3 (as there are 3 consecutive integers) > n^5n is divisible by 10*3=30. Sufficient.
Answer: B.
P.S. You have factored n^5n incorrectly and also your reasoning for n^2(n+1)(n1) 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: (n1)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



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Re: Is x divisible by 30? [#permalink]
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01 Feb 2012, 14:44



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Re: Is x divisible by 30? [#permalink]
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27 May 2013, 11:56
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 \(x = k*(m1)*m*(m+1)\). 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 =\(n^5n\) = \(n(n^21)(n^2+1)\) = \((n1)(n)(n+1)(n^2+1)\). 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 \((n^5n)\) as a product of 5 consecutive integers, then I would be succesfull in proving that \(n^5n\) is divisibly by 5 also. Now, assuming n as the median,to get a product of 5 consecutive integers, we would need (n2) and (n+2)>\((n^24)\)>\((n1)n(n+1)[(n^24)+(4+1)]\) = (n2)(n1)n(n+1)(n+2) + 5(n1)n(n+1)> This will be always be divisible by 30. There is actually a very famous theorem which states that \(x^px\) is always divisible by p, provided p is prime and x an integer. However, it is not relevant w.r.t GMAT.
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Re: Is x divisible by 30? [#permalink]
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04 Sep 2013, 12:10
Hey Bunuel
you are awesome!



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Re: Is x divisible by 30? [#permalink]
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30 Oct 2013, 09:30
Bunuel wrote: anuu wrote: Hi Bunnel,
I've a doubt in the following:
(n1)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^5n 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?



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Re: Is x divisible by 30? [#permalink]
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30 Oct 2013, 09:37
ronr34 wrote: Bunuel wrote: anuu wrote: Hi Bunnel,
I've a doubt in the following:
(n1)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^5n 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^5n is ALWAYS zero, so its ALWAYS divisible by 10. Next, n^5n=n(n^41)=n(n^21)(n^2+1)=(n1)n(n+1)(n^2+1) > (n1)n(n+1) is divisible by 3 (as there are 3 consecutive integers) > n^5n is divisible by 10*3=30. Sufficient.
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Re: Is x divisible by 30? [#permalink]
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08 Mar 2017, 13:21
Infer  we need 2*3*5 in the dividend. S1) k * m (m1) (m+1) if m is 10 sufficient but if m = 11 not sufficient. Also no information on k. K can be same as m. Not sufficient. S2) x = n (n1) (n+1) (n^2+1). Consider n as odd and even value. n = 12 then 11,12,13,145 sufficient n = 11 then 9,10,11,122 sufficient as we have 2*3*5
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Re: Is x divisible by 30? [#permalink]
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10 Mar 2017, 14:59
coolkl wrote: Infer  we need 2*3*5 in the dividend. S1) k * m (m1) (m+1) if m is 10 sufficient but if m = 11 not sufficient. Also no information on k. K can be same as m. Not sufficient.
S2) x = n (n1) (n+1) (n^2+1). Consider n as odd and even value. n = 12 then 11,12,13,145 sufficient n = 11 then 9,10,11,122 sufficient as we have 2*3*5 Great solution for S2. I'll just generalize it a bit further. For n=10, 11, 14 or 15 we already have a 5 factor in (n1)n(n+1). In fact, this applies to any n with units digit of 0, 1, 4, 5 as well as 6 (because 1+ 5) and 9 (because 4+ 5). There is only need for testing for n when n has 2 or 3 as a units digits* (12, 13, 22, 23, ...) because they don't have a 5 factor in (n1)n(n+1). The goal here is to find out if there is a factor 5 in (n²+1). The answer is YES, because if units digit of n=2, units digit of (n²+1)=5 units digit of n=3, units digit of (n²+1)=0 Therefore S2 is sufficient. *No need for testing unit digit 7 and 8 because they are just a factor 5 away from 2 and 3.



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Is x divisible by 30? [#permalink]
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07 Oct 2017, 10:12
Somehow, I came across to the following axiom: If p is a prime number, then for any integer n, \((n^pn)\) is ALWAYS divisible by both p and 3.If the above axiom is correct, we may solve the given problem in the following way. To be divisible by 30, the last digit of the number must be 0. (1) \(x = k*(m^3  m)\) From this we can say that \((m^3  m)\) is divisible by 3, but we can not say whether its last digit is 0. If k or \((m^3  m)\) is multiple of 10, then the answer is YES, otherwise the the answer is NO. Not sufficient. (2) \(x = n^5  n\) According to the above mentioned theory 1, \(n^5  n\) is divisible by both by 3 and 5. So, it is divisible by 15. Moreover, as Bunuel explained earlier, the last digit of \(n^5n\) is ALWAYS zero, so its ALWAYS divisible by 10. So, it must be divisible by 30 (LCM of 15 and 10). Sufficient. Answer is B.
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Re: Is x divisible by 30? [#permalink]
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30 Dec 2017, 10:27
why is the last digit of n^5−n ALWAYS zero ?



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