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If n is a positive integer, is 3^n+3^8 divisible by 4?

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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If n is a positive integer, is 3^n+3^8 divisible by 4?  [#permalink]

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18 Oct 2017, 03:15
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[GMAT math practice question]

If n is a positive integer, is $$3^n+3^8$$ divisible by 4?

1) n is a multiple of 3.
2) $$3^2 +3^{n+2}$$ is divisible by 4

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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" PS Forum Moderator Joined: 25 Feb 2013 Posts: 1203 Location: India GPA: 3.82 If n is a positive integer, is 3^n+3^8 divisible by 4? [#permalink] Show Tags 18 Oct 2017, 06:28 1 MathRevolution wrote: [GMAT math practice question] If n is a positive integer, is $$3^n+3^8$$ divisible by 4? 1) n is a multiple of 3. 2) $$3^2 +3^{n+2}$$ is divisible by 4 Statement 1: if $$n=3$$, then $$3^n+3^8$$ = $$3^3+3^8$$ => $$3^3(1+3^5)$$ => $$3^3*244$$, hence divisible by $$4$$ but if $$n=6$$, then $$3^n+3^8$$ = $$3^6+3^8$$ => $$3^6(1+3^2)$$ => $$3^6*10$$, hence not divisible by $$4$$. Hence Insufficient Statement 2: $$3^2 +3^{n+2}$$= $$3^2(1+3^n)$$, now this is a multiple of $$4$$, hence can be written as $$3^2(1+3^n)=4q$$, so $$3^n=\frac{4q}{9}-1$$, as $$3^n$$ is an integer so $$\frac{q}{9}$$ must be an integer, so let $$\frac{q}{9}=k$$ Hence $$3^n=4k-1$$, substitute this in question stem to get $$4k-1+3^8 = 4k+(3^4-1)*(3^4+1)=4k+80*82$$. Hence divisible by $$4$$. Sufficient Option B Manager Joined: 21 Jun 2014 Posts: 65 Re: If n is a positive integer, is 3^n+3^8 divisible by 4? [#permalink] Show Tags 18 Oct 2017, 10:40 The question can be simply narrowed down to " is n odd?" (3 ^n)/4 leaves remainder 3 if n is odd and it leaves remainder 1 of n is even. As 3^8 is leaving remainder 1, we need 3^n to leave remainder 3 and hence n has to be odd. From statement 1 n can be either odd or even so insuff. Statement 2 : we can infer that n+2 is odd and hence n is odd sufficient. OA: B Sent from my Moto G (5) Plus using GMAT Club Forum mobile app Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6028 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If n is a positive integer, is 3^n+3^8 divisible by 4? [#permalink] Show Tags 20 Oct 2017, 01:07 => Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures a solution. The first step of VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations. Since 38 has a remainder 1 when it is divided by 4, the question asks if 3n has a remainder 3 if it is divided by 4. 3^1 = 3 has the remainder 3, when it is divided by 4. 3^2 = 9 has the remainder 1, when it is divided by 4. 3^3 = 27 has the remainder 3, when it is divided by 4. 3^4 = 81 has the remainder 1, when it is divided by 4. ... Thus, the question asks if n is an odd number. Condition 1) Since n is a multiple of 3, it is unknown if n is odd or even. This is not sufficient. Condition 2) “3^2 +3^n+2 is divisible by 4” means 3^n+2 has a remainder of 3 when divided by 4. It means n + 2 and n are odd integers. Thus, this condition is sufficient. Therefore, the answer is B. Answer: B _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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If n is a positive integer, is 3^n+3^8 divisible by 4?  [#permalink]

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22 Oct 2017, 12:09
MathRevolution wrote:
[GMAT math practice question]

If n is a positive integer, is $$3^n+3^8$$ divisible by 4?

1) n is a multiple of 3.
2) $$3^2 +3^{n+2}$$ is divisible by 4

I notice we can apply the Binomial Theorem which is fast and convenient most of the time.
The Binomial Theorem states that, where n is a positive integer:
(a + b)n = an + (nC1)an-1b + (nC2)an-2b2 + … + (nCn-1)abn-1 + bn

With regards to our case :
$$3^n+3^8$$ = ( 4 - 1 )^n + (4 - 1) ^ 8 = 4^n + (nC1)*4^(n-1)*(-1)....+ (-1)^n + 4^8 + 8*4^7(-1) + ... + 8*4*(-1)^7 + (-1)^8

As the terms in Italic are divisible by 4, we only have to notice the remaining in Bold: (-1)^n + (-1)^8 = (-1)^n + 1
Case 1: If n is even, then the sum of remainder = 1 + 1 = 2.
Case 2: If n is odd, then the sum of remainder = -1 + 1 = 0

1) n = 3k
If k is even, then n is even.
If k is odd, then n is odd.
Insufficient.

2) $$3^2 +3^{n+2}$$ is divisible by 4[/quote]
Again 3^(n+2) +3^2 = [(4 - 1)^(n+2) = 4^(n+2) + ... + (-1)^(n+2)] + (4*2 +1)
The remainder would be (-1)^(n+2) + 1 = 0
=> (-1)^(n+2) = -1
=> (n+2) is odd
=> n is odd => Case 2.
So B is sufficient.
If n is a positive integer, is 3^n+3^8 divisible by 4? &nbs [#permalink] 22 Oct 2017, 12:09
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