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If n is a positive integer, what is the remainder when

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If n is a positive integer, what is the remainder when [#permalink] New post 19 Nov 2010, 08:40
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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
[Reveal] Spoiler: OA

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Re: Exponents [#permalink] New post 19 Nov 2010, 08:49
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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


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.

Answer: E.

Check Number theory chapter of Math Book for more: math-number-theory-88376.html

Hope it helps.
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Re: Exponents [#permalink] New post 19 Nov 2010, 08:59
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.
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Re: Exponents [#permalink] New post 19 Nov 2010, 10:42
Great explanation.
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Re: Exponents [#permalink] New post 20 Nov 2010, 02:17
put n=0, then we have (3^3)+2=29. now 29/5, we have r=4. is my short cut ok ???

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Re: Exponents [#permalink] New post 20 Nov 2010, 12:20
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anandthiru wrote:
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.
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Re: Exponents [#permalink] New post 22 Nov 2010, 13:38
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
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Re: If n is a positive integer, what is the remainder when [#permalink] New post 06 Jun 2013, 05:33
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Re: If n is a positive integer, what is the remainder when [#permalink] New post 06 Jun 2013, 14:00
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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
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Re: Exponents [#permalink] New post 06 Jun 2013, 14:20
Bunuel wrote:
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


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.

Answer: E.

Check Number theory chapter of Math Book for more: math-number-theory-88376.html

Hope it helps.

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
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Re: Exponents [#permalink] New post 06 Jun 2013, 15:10
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pavan2185 wrote:
Bunuel wrote:
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


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.

Answer: E.

Check Number theory chapter of Math Book for more: math-number-theory-88376.html

Hope it helps.

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.

Kudos if that helps. :-D
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Re: If n is a positive integer, what is the remainder when [#permalink] New post 07 Jun 2013, 01:11
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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.
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Re: If n is a positive integer, what is the remainder when [#permalink] New post 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

Hope its clear
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Re: If n is a positive integer, what is the remainder when [#permalink] New post 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.
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Re: If n is a positive integer, what is the remainder when [#permalink] New post 11 Apr 2014, 05:06
[3^(8n+3) + 2]/5 --> Calc. Remainder =>

within 30 sec, check for n=1,=2,=3

numerator becomes
(3^11+2),( 3^19+2),(3^27+2)

3 has cyclicity of 4 =>
Remainder from 11/4, 19/4, 27/4 = 3

from which last digit happens to be 7

therefore, we have drilled down the question to calculate remainder of (7+2)/5 i.e. 4 i.e. option E
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Remainders and Divisibility [#permalink] New post 28 Jul 2014, 17:12
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
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Re: Remainders and Divisibility [#permalink] New post 28 Jul 2014, 17:22
suhaschan 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


Get the cyclicity of 3 to get the units digit of the power of 3.

Since n is a positive integer, Lets try picking a number

n=1, 3^11

Cyclicity of 3 is 4 (3,9,27,81,243, repeat). Therefore

n=1, 3^11 --> units digit is 7

7+2=9/5= Remainder of 4.

Answer is E
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Re: Remainders and Divisibility [#permalink] New post 28 Jul 2014, 20:54
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suhaschan 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


You should know the concept of cyclicity. Powers of 3 have a cyclicity of 4 i.e.

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
3^6 = 729

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.
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Re: If n is a positive integer, what is the remainder when [#permalink] New post 29 Jul 2014, 00:36
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suhaschan 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


Merging topics. Please refer to the discussion above.

Also, please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 1 and 3. Thank you.


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Re: If n is a positive integer, what is the remainder when [#permalink] New post 29 Jul 2014, 10:21
Thanks a lot for the patient and fantastic explanation.
Re: If n is a positive integer, what is the remainder when   [#permalink] 29 Jul 2014, 10:21
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