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by Stuart@KaplanGMAT » Thu Mar 11, 2010 2:31 pm
roger federer wrote:If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?
A. 0
B. 1
C. 2
D. 3
E. 4
Let's start by analyzing the problem:

we've got some gigantic number and we're asked to find the remainder when we divide by 5. On the GMAT, will we ever have to calculate gigantic numbers? NO! So, there must be another way to solve.

One key thing to note is that we're dividing by 5; this makes us very happy! Why? Because to find the remainder when you divide by 5, you only need the last digit of the number.

So, we think: to solve the question, we just need to figure out the last digit of the gigantic number.

We then see that we're dealing with powers of 3. Let's jot some down on our scrap paper and look for a pattern:

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = ...3
3^6 = ...9

at this point we see that we have a 4 step repeating pattern: the units digit will cycle through 3, 9, 7 and 1.

Now that we have our pattern, let's apply it to the question at hand:

3^(8n+3) + 2

For simplicity, let's let n=1. That gives us:

3^11 + 2

Working through our repeating pattern, we see that 11 is going to be the 3rd step and that the units digit of 3^11 is 7.

Accordingly, the units digit of 3^11 + 2 will be 7 + 2 = 9.

Finally, when we divide 9 by 5, we have a remainder of 4... choose (E).
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by thebigkats » Sun Feb 13, 2011 11:50 am
Hi:

3^(8n+3)+2
= (3^(8n) * 3^3) + 2
= 3^8n * 27 + 2
= 27 * (3^8)^n + 2
= 27 * (3*3*3*3*3*3*3*3) ^n + 2
= 27 * (81 * 81)^n + 2

Now we know that 81*81= xx..xx1 (i.w. units digit is always 1)
We also know that when you multiple a no with unit digit 1 with another no with units digit 1, the end result would also have units digit of 1
So (81*81)^n has unit digit of 1
Multiplying this no with 27 will give us a no with unit digit of 7 (7*1)
adding 2 to this no will give us units digit of 9

and remainder = 4
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by sushantgupta » Sat Jul 16, 2011 9:53 pm
lets rewrite the equation

((3^8n)*(25+2)+2 )
3^8n*25 is divisible by 5 so we can drop it from further calculation

(3^8n)*2 + 2 = (9*9*9*9)*2+2 = (81*81)*2 +2 last digit of 81*81 = 1
so remainder = 1*2+2 = 4 hence E
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by mirantdon » Sun Jul 17, 2011 2:04 am
+1 for E .
Power series of 3 .
(3 9 7 1 )
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by ritzzzr » Mon Jan 23, 2012 12:09 am
4 must be the remainder!!
the power cycle of 3 is 3,9,7,1
and since 3^(8n+3) will have always last digit as 7 and by adding two last digit will be always 9,so by dividing the number by 5,we will have 4 as remainder
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by ronnie1985 » Wed Apr 04, 2012 8:51 am
Using remainder theorem,

R[{3^(8n+3)+2}/5] = R[{27*3^8n+2}/5] = R[{2*9^4n+2}/5] = R[{2*(-1)^4n+2}/5] = R[{2+2}/5] = 4
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by anujan007 » Fri Jul 20, 2012 10:14 am
Plugged in few numbers for n to find the solution. We need to find the units digit of the mentioned value to identify the remainder when divided by 5.

Substituted n=1, (3)^11 = (3)^8*(3)^3 = (81)^2*27 = XXX7 + 2 = XXX9, when divided by 5, remainder =4

Substituted n=2, (3)^19 = (3)^16*(3)^3 = (81)^4*27 = XXX7 + 2 = XXX9, when divided by 5, remainder =4

hence chose the answer as E.
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by mariofelixpasku » Thu Mar 07, 2013 1:40 pm
simplest way as it doesnt say cannot be determined, insert n=0 and you get the remainder = 4
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by rajeshsinghgmat » Sat Apr 20, 2013 12:46 am
(C) 2

k=8n+3

3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243, 3^6=729, 3^7=2187
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by jaspreetsra » Thu Jan 01, 2015 12:33 am
IMO: E
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by Matt@VeritasPrep » Mon Jan 05, 2015 11:14 am
The answer is definitely E, as Stuart demonstrated quite clearly five years ago - not sure why we have so many replies here!

Another approach, if there's any confusion:

3���³ + 2 is really
3�� * 3³ + 2 is really
(3�)� * 27 + 2 is really
(6561)� * 27 + 2 is really
(...1) * 27 + 2 is really
(...7) + 2 is really
(...9)

Anything that ends in 9 has a remainder of 4 when divided by 5, so we're done.

(FYI: The ...1 indicates "a really big number that ends in 1".)
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