Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

It should be 3^{8n+3}+2.

The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit: 3^1 --> the units digit is 3; 3^2 --> the units digit is 9; 3^3 --> the units digit is 7; 3^4 --> the units digit is 1; 3^5 --> the units digit is 3 AGAIN; ...

So, the units digit repeats the following pattern {3-9-7-1}-{3-9-7-1}-.... 3^{8n+3} will have the same units digit as 3^3, which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of 3^{8n+3}+2 will be 7+2=9. Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4.

ahhh okay. i followed the same pattern for the exponents. the problem made it seem like +2 was in the exponent. makes perfect sense now. much appreciated, Bunuel. _________________

put n=0, then we have (3^3)+2=29. now 29/5, we have r=4. is my short cut ok ???

counter arguments welcome instead of kudos !!!

n is given to be a positive integer so it cannot be 0. That said, whatever will be the last digit of 3^{(8n+3)} + 2 for n = positive integer, will be the last digit when n = 0. (We know cyclicity works for n = 0 too) Hence your short cut works. _________________

Answer should be E i.e. 4 remember 3 has a cyclicity of 4 =>3,9,*7,*1 3^(8n+3) +2 => for n=1 => 8*1+3 =>11 =>3(on dividing 11/4) for n=2 =>19=>3(on dividing 19/4) for n=3 =>27=>3(on dividing 27/4) for n=4 =>35=>3(on dividing 35/4) for n=5 =>43=>3(on dividing 43/4) for n=6 =>51=>3(on dividing 51/4) therefore 3^3 gives 7 in units place => 7+2 => 9 in all the cases hence on dividing 9/5 remainder is 4 _________________

I will give a Fight till the End

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed." - Bernard Edmonds

A person who is afraid of Failure can never succeed -- Amneet Padda

Re: If n is a positive integer, what is the remainder when [#permalink]
06 Jun 2013, 14:00

2

This post received KUDOS

you can split it up to make it look simpler. if x = 1 (3^8)(3^3) + 2 3^8 = 9^4 = 9 x 9 x 9 x 9 = 81 x 81 (units digit is 1) 3^3 = 27 (units digit is 7) 1 x 7 + 2 = 9 9/5 = 1 r4 Answer is E

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

It should be 3^{8n+3}+2.

The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit: 3^1 --> the units digit is 3; 3^2 --> the units digit is 9; 3^3 --> the units digit is 7; 3^4 --> the units digit is 1; 3^5 --> the units digit is 3 AGAIN; ...

So, the units digit repeats the following pattern {3-9-7-1}-{3-9-7-1}-.... 3^{8n+3} will have the same units digit as 3^3, which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of 3^{8n+3}+2 will be 7+2=9. Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4.

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

It should be 3^{8n+3}+2.

The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit: 3^1 --> the units digit is 3; 3^2 --> the units digit is 9; 3^3 --> the units digit is 7; 3^4 --> the units digit is 1; 3^5 --> the units digit is 3 AGAIN; ...

So, the units digit repeats the following pattern {3-9-7-1}-{3-9-7-1}-.... 3^{8n+3} will have the same units digit as 3^3, which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of 3^{8n+3}+2 will be 7+2=9. Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4.

I just came to testing the cycle and felt a dead end. I am not getting why are we zeroing in on 3^3 for the first part of the addition in the question

The reason we are focusing on 3^3 is because the repeating cycle of 3^{8n} only matters because it is a sequence that loops in a pattern of 4 (which is consequently a divisor of 8).

You said that you tested the cycle. So you should understand that, as Bunuel said, 3-9-7-1-3-9-7-1 is the cycle. No matter what multiple of 8 you use (8, 16, 24, 32, etc.), 1 will always be the units digit of the final value. If that is the case we can actually forget the 8n part of the equation, because you know that 3^{8n} will always have a units digit of 1. This means that you have to focus on the 3^3 instead. You can look at this is two ways. Either go three more steps in the cycle and land on 7 (3-9-7), or because 3^{8n+3} = 3^{8n}*3^3 = (number with the units digit of 1) x 27. The units digit will be 7 because 1 x 7 is 7. 7+2 = 9, 9/5 = 1 r4. Answer is E.

You can also look at my post above for a method that doesn't involve all this sequencing stuff

On another note, be careful not to make this the sequencing a rule for all questions that look like this because it definitely is not a rule. If you see 3^{7m+3}, you can't just "cancel out" the 7m. If m=1 the cycle is 3-9-7-1-3-9-7, but if m=2 it continues the sequence and does not restart back at 3 but instead starts where it left off after 7 (1-3-9-7-1-3-9).

The reason we focus on the 3^3 in this equation is only because the final value of the "8n sequence" always ends with the same units digit.

Re: If n is a positive integer, what is the remainder when [#permalink]
07 Jun 2013, 01:11

2

This post received KUDOS

Expert's post

azule45 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

We know that Remainder(\frac{a*b}{c}) = Remainder(\frac{a}{c})*Remainder(\frac{b}{c}) for integral values of a,b and c.

The given expression : \frac{(3^{8n}*3^3 + 2)}{5} = \frac{27*(3^4)^{2n}}{5}+\frac{2}{5} = Remainder of \frac{(27)*(81)^{2n}}{5} + Remainder of \frac{2}{5} = Remainder of\frac{(2)*(1)^{2n}}{5}+Remainder of \frac{2}{5}= Remainder of \frac{4}{5} = 4. _________________

Re: If n is a positive integer, what is the remainder when [#permalink]
03 Feb 2014, 09:32

azule45 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

I followed a slightly longer approach

Assume n=1 then 3^11 + 2 /5, what is the remainder?

Well 3^11 is (5-2)^11 so we have -2^11/5 what is the remainder of this?

Again -2*4^5 = -2*(5-1)^5, which gives a remainder of -1 multiplied by -2 gives a remainder of 2, and if we add the 2 that was missing for the first part of the question we get 4 as a remainder

Re: If n is a positive integer, what is the remainder when [#permalink]
09 Apr 2014, 07:03

\frac{(3^8)^n*3^3 + 2}{5}

\frac{81^n * 81^n * 27 + 2}{5} Making use of binomial theorem:

\frac{ (80+1)^n * (80+1)^n * 27 + 2 }{5}

\frac{ 1^n * 1^n * 2 + 2 }{5}

\frac{2+2}{5}

\frac{4}{5}

Thus 4 is remainder. _________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Look at the last digits of powers: 3, 9, 7, 1, 3, 9, 7, 1, 3, 9...

So whenever we have a power of 3, we remove groups of 4 out of it i.e. divide it by 4 and whatever is the remainder tells us the last digit.

So say we have 3^13, we divide 13 by 4 and are left with 1 remainder. So this is the first term of the next cycle and will end with 3. Say we have 3^39, we divide 39 by 4 and are left with 3 remainder. So this is the third term of the next cycle and will end with 7.

If you have 3^{8n+3}, we divide 8n+3 by 4 and are left with 3 remainder. So this is the third term of the next cycle and will end with 7.

Last term of 3^{8n+3}+2 will be 7+2 = 9. When you divide a number that ends with 9 by 5, you get 4 remainder. If you are not sure why this is so, check out the first part of the question discussed in this post: http://www.veritasprep.com/blog/2014/03 ... emainders/

Answer (E)

Or since you know that cyclicity of 3 is 4, just put n = 0 to get 3^3 + 2 = 29. When you divide 29 by 5, you get remainder 4. _________________