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If n is an integer greater than 1 is 3^n2^n divisible by 35
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Updated on: 22 Jul 2013, 05:17
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Originally posted by Bunuel on 08 Oct 2009, 06:20.
Last edited by Bunuel on 22 Jul 2013, 05:17, edited 2 times in total.
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Re: Divisibility
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Updated on: 07 Nov 2009, 08:45
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?
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Originally posted by Bunuel on 08 Oct 2009, 11:26.
Last edited by Bunuel on 07 Nov 2009, 08:45, edited 1 time in total.




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Re: Divisibility
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08 Oct 2009, 10:55
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.
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Re: Divisibility
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08 Oct 2009, 11:40
well I have been preparing for only a month now so am probably not the best person to answer this. Going by the problems I have found on this forum it was slightly on the tougher side. But then again wont do anyone any harm to remember one simple rule
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Re: Divisibility
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09 Oct 2009, 04:26
Very very good question U should give more time for the others



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Re: Divisibility
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09 Oct 2009, 19:03
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. +1 for both of you for 750+ level question and solution...
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Re: Divisibility
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09 Oct 2009, 21:12



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Re: Divisibility
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09 Oct 2009, 22:46
Bunuel wrote: Thanks GMAT TIGER, but frankly speaking when I was composing this question I didn't think that it would be ranked as 750+.
Is it because the rule needed to deal with this problem is not widely known? The question is perfect. It looks tough but solution is simple if you persue the right approach. Thats the characterstics of 750+ question, which are not essentially hard/difficult.....
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Re: Divisibility
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09 May 2010, 03:03
Great question! I hope you can add more of these! I've seen a few of these questions but at lower level of difficulty. Usually the divisor is a factor of (ab) or (a + b) where a, b are the base. So, it is easier to relate. Having 35 as divisor stumped me as I couldn't relate it to 3 and 2. I noticed that there are not a lot of discussion about this type of question either on the net or in the gmat book. So, I'm really glad to find it! Hope to see more!!!!Kudos for all your efforts. Cant tell you how much it helps..since I dont have access to any gmat books.



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Re: Divisibility
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16 May 2010, 11:43
its a indeed a great quality question.



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Re: Divisibility
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09 Dec 2010, 10:45
Bunuel wrote: [ 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.
What is the reasoning behind these rules? (understanding makes it easier to remember)



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Re: Divisibility
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09 Dec 2010, 12:24
Fijisurf wrote: Bunuel wrote: [ 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.
What is the reasoning behind these rules? (understanding makes it easier to remember) I really doubt you need this. RULE: for \(x^ny^n\):\(x^ny^n=(xy)(x^{n1}+x^{n2}y+x^{n3}y^2+x^{n4}y^3+...+xy^{n2}+y^{n1})\) So, \(x^ny^n\) is always divisible by \(xy\). Now, when \(n=even\), then the second multiple will have n, so even terms and \((x^{n1}+x^{n2}y+x^{n3}y^2+x^{n4}y^3+...+xy^{n2}+y^{n1})=\) \(=(x^{n2}(x+y)+x^{n4}y^2(x+y)+...+y^{n2}(x+y))=(x+y)(x^{n2}+x^{n4}y^2+...+y^{n2})\) So, \(x^ny^n\) is also divisible by \(x+y\) when \(n\) is even. Consider the following examples: \(x^2y^2=(xy)(x+y)\): divisible by both \(xy\) and \(x+y\); \(x^3y^3=(xy)(x^2+xy+y^2)\): divisible by \(xy\). RULE: for \(x^n+y^n\):When \(n=odd\) then: \(x^n+y^n=(x+y)(x^{n1}x^{n2}y+x^{n3}y^2x^{n4}y^3+...xy^{n2}+y^{n1})\) So, when \(n=odd\) then \(x^n+y^n\) is divisible by \(x+y\). When \(n=even\) then \(x^n+y^n\) is NOT divisible by either \(x+y\) or \(xy\). Consider the following example: \(x^3+y^3=(x+y)(x^2xy+y^2)\): divisible by \(x+y\).
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Re: Divisibility
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09 Dec 2010, 13:32
Thanks! I hope this does not appear on GMAT too often.



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05 Aug 2011, 23:11
hmmm. this is the kind of questions which make the prep interesting for anyone.



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Re: If n is an integer >1, is 3^n2^n divisible by 35?
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07 Jul 2013, 23:54



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08 Jul 2013, 02:32
Fijisurf wrote: Bunuel wrote: [ 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.
What is the reasoning behind these rules? (understanding makes it easier to remember) Bunuel wrote: 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.
Explanation to follow. I just wanted to show another method to prove the above. Perfect solution is already given by everyone. +1 for the sum. For any function f(x), if x=a makes f(a)=0, then (xa) is a factor of the given function. This is something we all know. It has just been put in words by me.To illustrate that, consider the given example : f(x) =\(x^22x+1\) ; for x=1, f(1) = 0. Thus, (x1) is a factor of f(x),and as everyone would have recognised,we all know f(x) = \((x1)^2\). I.Function of the form f(x) = \(x^n+y^n\) will have (x+y) as a factor, if and only if for x=y, f(y) = 0. Thus, replacing the value of x in the above function, we get : \((y)^n+y^n \to\) Notice that this will be zero , only if n is odd. \(x^n+y^n\) will always have (x+y) as a factor, if n = odd.II. Function of the form f(x) = \(x^ny^n\). Just as above, for (x+y) to be a factor, for x = y,f(y) = 0 Again, by replacing f(y) = \((y)^ny^n \to\) This will be zero only if n is even. \(x^ny^n\) will always have (x+y) as a factor, if n = even.Again, for (xy) to be a factor, for x = y, f(y) = 0. f(y) = \(y^ny^n\) = 0. \(x^ny^n\) will always have (xy) as a factor, if n = odd/even.
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Re: If n is an integer >1, is 3^n2^n divisible by 35?
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08 Jul 2013, 13:08
very good question... thanks...



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Re: Divisibility
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23 Jul 2013, 03:25
Bunuel wrote: Fijisurf wrote: Bunuel wrote: [ 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.
What is the reasoning behind these rules? (understanding makes it easier to remember) I really doubt you need this. RULE: for \(x^ny^n\):\(x^ny^n=(xy)(x^{n1}+x^{n2}y+x^{n3}y^2+x^{n4}y^3+...+xy^{n2}+y^{n1})\) So, \(x^ny^n\) is always divisible by \(xy\). Now, when \(n=even\), then the second multiple will have n, so even terms and \((x^{n1}+x^{n2}y+x^{n3}y^2+x^{n4}y^3+...+xy^{n2}+y^{n1})=\) \(=(x^{n2}(x+y)+x^{n4}y^2(x+y)+...+y^{n2}(x+y))=(x+y)(x^{n2}+x^{n4}y^2+...+y^{n2})\) So, \(x^ny^n\) is also divisible by \(xy\) when \(n\) is even. Consider the following examples: \(x^2y^2=(xy)(x+y)\): divisible by both \(xy\) and \(x+y\); \(x^3y^3=(xy)(x^2+xy+y^2)\): divisible by \(xy\). RULE: for \(x^n+y^n\):When \(n=odd\) then: \(x^n+y^n=(x+y)(x^{n1}x^{n2}y+x^{n3}y^2x^{n4}y^3+...xy^{n2}+y^{n1})\) So, when \(n=odd\) then \(x^n+y^n\) is divisible by \(x+y\). When \(n=even\) then \(x^n+y^n\) is NOT divisible by either \(x+y\) or \(xy\). Consider the following example: \(x^3+y^3=(x+y)(x^2xy+y^2)\): divisible by \(x+y\). Good rules to remember but I think there is a typo above ,When you are explaining the logic for even case I think you meant to write So, \(x^ny^n\) is also divisible by \(x+y\) when \(n\) is even and not \(xy\). Good rules to keep in mind.



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Re: Divisibility
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Re: Divisibility &nbs
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