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Re: Divisibility
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27 Dec 2013, 07:44
Bunuel wrote: jax91 wrote: Bunuel wrote: Let's try one more.
If n is an integer >1, is 3^n2^n divisible by 35?
(1) n is divisible by 15. (2) n is divisible by 18.
How you like this one?
Explanation to follow. we need to split 35 35 = 27 + 8 (1) 3^15n2^15n = (3^3)^5n  (2^3)^5n = 27^5n  8^5n considering n to be 1 (odd) we get that 27^5 ends with a 7, and 8^5 ends with an 8. so the last digit will be 9 . 35 cant divide a number ending with 9 consider n to be even now, say 2 i.e. 27^10  8^10 . now we will need some corollaries we get from the binomail theorem. if we have a number as a^n  b^n then if n is odd , the number is divisible by a  b if n is even, the number is divisible by a + b so this makes 27^108^10 divisible by 27 + 8 = 35. thus an even value of n will make the term 3^15n  2^15n divisible by 35, while an odd value will not be divisble So 1 is insufficient. By using the above mentioned rule we can infer that 3^18n2^18n (27^6n  8 ^6n) will always be divisible by 35 as the index will always be even. So 2 is sufficient . So B. Exactly. +1 to you. RULE: for x^ny^n: \(x^ny^n\) is ALWAYS divisible by \(xy\). \(x^ny^n\) is divisible by \(x+y\) when n is even. If n is an integer >1, is \(3^n2^n\) divisible by 35? (1) n is divisible by 15. > \(3^{15m}2^{15m}=27^{5m}8^{5m}\) > 5m may or my not be even, so insufficient to answer, whether it's divisible by 27+8=35. (2) n is divisible by 18. > \(3^{18m}2^{18m}=27^{6m}8^{6m}\) > 6m is even, so 3^18m2^18m=27^6m8^6m is divisible by 27+8=35. Sufficient. B. jax91 can you please tell me whether this question was too hard for GMAT? Yes, this actually ranks as top 50 of the hardest questions I've done on GMAT



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If n is an integer greater than 1 is 3^n2^n divisible by 35
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02 Nov 2015, 02:59
Bunuel wrote: If n is an integer greater than 1, is 3^n2^n divisible by 35?
(1) n is divisible by 15. (2) n is divisible by 18.
Explanation to follow. I didnt know the rule, but i solved it in the following way: It is said that: 3^n2^n=35a (1) n=15b 3 has a repeated last digits when it is raised to the power 3, 9, 7, 1 2 has a repeated last digits when it is raised to the power 2, 4, 8, 6 if n = 15b > > the last digit of 3^15b2^15b = ...7  ...8 = ....9 9 is not divisible by 35 INSUFFICIENT (2) n=18c 3 has a repeated last digits when it is raised to the power 3, 9, 7, 1 2 has a repeated last digits when it is raised to the power 2, 4, 8, 6 if n = 18c > > the last digit of 3^18c2^18c = ...9  ...4 = ....5 5 is divisible by 35 SUFFICIENT Answer B I know, that it is more like intuition than solution))



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Re: If n is an integer greater than 1 is 3^n2^n divisible by 35
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14 Nov 2015, 20:07
Bunuel wrote: muralimba wrote: Bunuel, Not to find out the mistake, but please correct the above typo. Shudn't that be \(x^3+y^3\) Thanks. Corrected the typo. Hey Bunuel, Where are all these unique rules and properties located? By doing the questions I keep learning new rules while I do the questions. Is there somewhere that sums these tricks up?



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Re: If n is an integer greater than 1 is 3^n2^n divisible by 35
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15 Nov 2015, 08:13



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Re: If n is an integer greater than 1 is 3^n2^n divisible by 35
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24 Aug 2016, 05:28
Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HEREAnother solution maybe ?? Cyclicity of 3 => 3,9,7,1 Cyclicity of 2 => 2,4,8,6 Now as n is same => to be divisible by 35 => it must be divisible by 5 only way that is possible is if N=4K+2 for some K Statement 1 => N=15*P => N can be 15 => No N can be 30 > Yes => Not suff Statement 2 => N is 18*P => 18=(4k+2) Hence it will work => Suff SMASH THAT B
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If n is an integer greater than 1 is 3^n2^n divisible by 35
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07 Jul 2017, 09:45
olnata wrote: Bunuel wrote: If n is an integer greater than 1, is 3^n2^n divisible by 35?
(1) n is divisible by 15. (2) n is divisible by 18.
Explanation to follow. I didnt know the rule, but i solved it in the following way: It is said that: 3^n2^n=35a (1) n=15b 3 has a repeated last digits when it is raised to the power 3, 9, 7, 1 2 has a repeated last digits when it is raised to the power 2, 4, 8, 6 if n = 15b > > the last digit of 3^15b2^15b = ...7  ...8 = ....9 9 is not divisible by 35 INSUFFICIENT (2) n=18c 3 has a repeated last digits when it is raised to the power 3, 9, 7, 1 2 has a repeated last digits when it is raised to the power 2, 4, 8, 6 if n = 18c > > the last digit of 3^18c2^18c = ...9  ...4 = ....5 5 is divisible by 35SUFFICIENT Answer B I know, that it is more like intuition than solution)) Hi Olnata! Although it is a good approach it is not fully correct. 35 (and any of its multiples) is divisible by 5, but not all multiples of 5 are divisible by 35. The multiple of 5 should also be a multiple of 7. This was indeed a very difficult question. Thanks Bunuel



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Re: If n is an integer greater than 1 is 3^n2^n divisible by 35
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10 Nov 2018, 10:16
GMATinsight is it necessary to use the rule mentioned in this post: x^n  y^n is always divisible by xy, and with x+y if n is even Is there any simpler way to solve this? Posted from my mobile device



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Re: If n is an integer greater than 1 is 3^n2^n divisible by 35
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10 Nov 2018, 20:49
s111 wrote: GMATinsight is it necessary to use the rule mentioned in this post: x^n  y^n is always divisible by xy, and with x+y if n is even Is there any simpler way to solve this? Posted from my mobile device Yes this rule is always correct. Using induction method you can make this observation \((x^1y^1)\) is divisible by \((xy)\) \((x^2y^2) = (x+y)(xy)\) is divisible by \((x+y)\) \((x^3y^3) = (xy)(x^2+y^2xy)\) is divisible by \((xy)\) \((x^4y^4) = (x^2y^2)*(x^2+y^2) = (x+y)(xy)*(x^2+y^2)\) is divisible by \((x+y)\) and so on...
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Re: If n is an integer greater than 1 is 3^n2^n divisible by 35
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10 Nov 2018, 23:49
Bunuel wrote: If n is an integer greater than 1, is 3^n2^n divisible by 35?
(1) n is divisible by 15. (2) n is divisible by 18.
Explanation to follow. This particular question of defining whether the expression would be an integer value or not can be solved by using the cyclicity rule: first factorize 35: 7*5 , and we know the unit cyclicity of 3 & 2 is 4; viz 3^1:3, 3^2:9: 3^3: 7: 3^4:1, 3^5:3 and for 2 ; 2^1:2 , 2^2:4, 2^3: 8, 2^4: 6,2^5:2.. from stmt 1: we can say that n can be 15,30,45,60 .. therefore 3^n2^n to be divisible by 7*5 it has to have a unit digit either 5 or 0, which is a yes and no case , so it is insufficient for stmt 1: n would be 18,36,54.. and for all values we get unit digits as divisible by 5, hence yes and sufficient .. IMO B is correct...




Re: If n is an integer greater than 1 is 3^n2^n divisible by 35 &nbs
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