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If n is an integer greater than 1 is 3^n-2^n divisible by 35

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Re: Divisibility  [#permalink]

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New post 27 Dec 2013, 08:44
Bunuel wrote:
jax91 wrote:
Bunuel wrote:
Let's try one more.

If n is an integer >1, is 3^n-2^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) 3^15n-2^15n = (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 27^10-8^10 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 3^18n-2^18n (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^n-y^n:
\(x^n-y^n\) is ALWAYS divisible by \(x-y\).
\(x^n-y^n\) is divisible by \(x+y\) when n is even.


If n is an integer >1, is \(3^n-2^n\) 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 3^18m-2^18m=27^6m-8^6m 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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Re: If n is an integer greater than 1 is 3^n-2^n divisible by 35  [#permalink]

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New post 14 Nov 2015, 21:07
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^n-2^n divisible by 35  [#permalink]

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New post 15 Nov 2015, 09:13
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GMATDemiGod wrote:
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?


Please check this post ALL YOU NEED FOR QUANT ! ! ! and links in my signature.
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Re: If n is an integer greater than 1 is 3^n-2^n divisible by 35  [#permalink]

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New post 24 Aug 2016, 06:28
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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To find DS questions by Kudos, sort by Kudos here: gmat-data-sufficiency-ds-141/
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Another 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^n-2^n divisible by 35  [#permalink]

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New post 07 Jul 2017, 10:45
olnata wrote:
Bunuel wrote:
If n is an integer greater than 1, is 3^n-2^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^n-2^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 3^15b-2^15b = ...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 3^18c-2^18c = ...9 - ...4 = ....5
5 is divisible by 35
SUFFICIENT

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^n-2^n divisible by 35  [#permalink]

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New post 16 Jul 2018, 03:29
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If n is an integer greater than 1 is 3^n-2^n divisible by 35 &nbs [#permalink] 16 Jul 2018, 03:29

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