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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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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]
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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