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If n is an integer greater than 1 is 3^n-2^n divisible by 35 27 Dec 2013, 08:44
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Bunuel
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Bunuel
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?
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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 02 Nov 2015, 03:59
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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.
Show more

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))
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If n is an integer greater than 1 is 3^n-2^n divisible by 35 14 Nov 2015, 21:07
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Bunuel
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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.
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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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If n is an integer greater than 1 is 3^n-2^n divisible by 35 15 Nov 2015, 09:13
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muralimba

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?
Show more

Please check this post ALL YOU NEED FOR QUANT ! ! ! and links in my signature.
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If n is an integer greater than 1 is 3^n-2^n divisible by 35 24 Aug 2016, 06:28
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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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 07 Jul 2017, 10:45
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olnata
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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))
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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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If n is an integer greater than 1 is 3^n-2^n divisible by 35 10 Nov 2018, 11:16
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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?

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If n is an integer greater than 1 is 3^n-2^n divisible by 35 10 Nov 2018, 21:49
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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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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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If n is an integer greater than 1 is 3^n-2^n divisible by 35 11 Nov 2018, 00:49
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Bunuel
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.
Show more

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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If n is an integer greater than 1 is 3^n-2^n divisible by 35 27 Mar 2025, 05:12
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