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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]

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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

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19 Nov 2010, 08:49
10
10
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.

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

Hope it helps.
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Re: If n is a positive integer, what is the remainder when  [#permalink]

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06 Jun 2013, 14:00
5
3
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
##### General Discussion
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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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19 Nov 2010, 10:42
Great explanation.
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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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20 Nov 2010, 12:20
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)
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22 Nov 2010, 13:38
2
1
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]

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06 Jun 2013, 05:33
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on remainders problems: remainders-144665.html

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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.

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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06 Jun 2013, 15:10
2
2
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.

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.
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Re: If n is a positive integer, what is the remainder when  [#permalink]

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07 Jun 2013, 01:11
2
1
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]

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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]

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09 Apr 2014, 07:03
1
$$\frac{(3^8)^n*3^3 + 2}{5}$$

$$\frac{81^n * 81^n * 27 + 2}{5}$$
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]

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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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28 Jul 2014, 17:22
1
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.

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28 Jul 2014, 20:54
2
1
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/

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]

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30 May 2015, 19:57
by the method of exponents break the exp as (3^8n * 3^3) +2
now , we know cyclicity of 3 is 4 that means unit digit is 1 when 3^4 so,3^8n =1 (bcz 8n is divisible by 4)
=> exp becomes 27*3^8n + 2 => 27 *xxxx1 (since n can be any integer)+2 => xxx7+2=> xxx9 and now xxx9/5 leaves rem(5) .

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Re: If n is a positive integer, what is the remainder when  [#permalink]

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09 Aug 2017, 11:51
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 need to determine the remainder of:

3^8n x 3^3 + 2 when it is divided by 5.

We only need to know the units digit of the above expression to determine the remainder.

Let’s look at the pattern of units digits of powers of 3. Note that we are only concerned with the units digits, so, for example, for 3^3, we concern ourselves only with the units digit of 27, which is 7. Here is the pattern:

3^1 = 3

3^2 = 9

3^3 = 7

3^4 = 1

3^5 = 3

The repeating pattern is 3-9-7-1. Since the units digit pattern for a base of 3 is 3-9-7-1, we see that whenever 3 is raised to an exponent that is a multiple of 4, the units digit will be 1. Thus:

3^8n x 3^3 = units digit of 1 x units digit of 7 = units digit of 7

So, units digit of 7 + 2 = units digit of 9, and thus 9/5 has a remainder of 4.

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Re: If n is a positive integer, what is the remainder when  [#permalink]

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19 Aug 2018, 23:56
2
concept of cyclicity
8n+3=4(2n)+3.
therefore the last digit will be 7.
7+2=9 hence when the number will be divided by 5, the remainder will be 4
Re: If n is a positive integer, what is the remainder when &nbs [#permalink] 19 Aug 2018, 23:56

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