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

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SVP
Joined: 06 Sep 2013
Posts: 1745
Concentration: Finance

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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^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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If n is an integer greater than 1 is 3^n-2^n divisible by 35  [#permalink]

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02 Nov 2015, 02:59
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

I know, that it is more like intuition than solution))
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Posts: 85
Concentration: General Management, Finance
GMAT 1: 680 Q46 V38
GMAT 2: 690 Q47 V38
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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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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?
Math Expert
Joined: 02 Sep 2009
Posts: 50613
Re: If n is an integer greater than 1 is 3^n-2^n divisible by 35  [#permalink]

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15 Nov 2015, 08:13
1
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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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 HERE

To find DS questions by Kudos, sort by Kudos here: gmat-data-sufficiency-ds-141/
To find PS questions by Kudos, sort by Kudos here: gmat-problem-solving-ps-140/

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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07 Jul 2017, 09:45
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

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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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 x-y, 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^n-2^n divisible by 35  [#permalink]

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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 x-y, 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

$$(x^1-y^1)$$ is divisible by $$(x-y)$$

$$(x^2-y^2) = (x+y)(x-y)$$ is divisible by $$(x+y)$$

$$(x^3-y^3) = (x-y)(x^2+y^2-xy)$$ is divisible by $$(x-y)$$

$$(x^4-y^4) = (x^2-y^2)*(x^2+y^2) = (x+y)(x-y)*(x^2+y^2)$$ is divisible by $$(x+y)$$

and so on...
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Re: If n is an integer greater than 1 is 3^n-2^n divisible by 35  [#permalink]

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10 Nov 2018, 23:49
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.

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^n-2^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^n-2^n divisible by 35 &nbs [#permalink] 10 Nov 2018, 23:49

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