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# What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6

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Manager
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What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6  [#permalink]

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06 Jul 2016, 08:07
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Question Stats:

65% (01:54) correct 35% (02:10) wrong based on 216 sessions

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What is the remainder when $$8^1+ 8^2+ 8^3……+8^{15}$$ is divided by 6

A. 0
B. 1
C. 2
D. 4
E. 5

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Re: What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6  [#permalink]

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06 Jul 2016, 23:52
1
C

8/6 has remendar 2 and 64/6 has remainder 4

so there will be 8 2's and 7 4's from the sum which when added give 44 as sum.So 44/6=2 remainder
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Re: What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6  [#permalink]

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07 Jul 2016, 08:14
hsbinfy wrote:
C

8/6 has remendar 2 and 64/6 has remainder 4

so there will be 8 2's and 7 4's from the sum which when added give 44 as sum.So 44/6=2 remainder

hsbinfy

I followed you till getting the sum as 44 (This is sum of individual remainders). But I could not follow the highlighted part. Can you please say why we need to divide the sum of the remainders again by 6?

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Re: What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6  [#permalink]

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09 Jul 2016, 10:24
2
Hello, I approached this question with using cyclisity properties, but i don't know whether it is correct. Maybe someone can prove confirm it mathematically.

So we need to determine the remainder of the sum of the factors of 8^1 till 8^15

I determined the cycle of 8 and got

8^1 = 8
8^2 = 64
8^3 = 512
8^4 = 4096
8^5 = 32768

So I just added all the ones digits we will get if we sum 8^1 to 8^15, so we would get [(8+4+2+6)*3 + 8+4+2]/6 = 74/6 = 12 with remainder = 2.

Can anyone tell me if it is correct or coincidence?
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Re: What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6  [#permalink]

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09 Jul 2016, 11:29
Matthias1205 wrote:
Hello, I approached this question with using cyclisity properties, but i don't know whether it is correct. Maybe someone can prove confirm it mathematically.

So we need to determine the remainder of the sum of the factors of 8^1 till 8^15

I determined the cycle of 8 and got

8^1 = 8
8^2 = 64
8^3 = 512
8^4 = 4096
8^5 = 32768

So I just added all the ones digits we will get if we sum 8^1 to 8^15, so we would get [(8+4+2+6)*3 + 8+4+2]/6 = 74/6 = 12 with remainder = 2.

Can anyone tell me if it is correct or coincidence?

No, it is not correct. See my post: http://gmatclub.com/forum/what-is-the-remainder-when-207423.html#p1591521
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Re: What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6  [#permalink]

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09 Nov 2016, 15:11
1
1
You'll notice a pattern with the remainder if you divide 8 by 6 first, 64/8 and so on....

Remainder will jump back and forth between 2 (on odd exponents) and 4 (on even exponents)

If you count the number of remainder 2's and 4's --> 8 2's & 7 4's--> 2(8)+7(4) = 16+28 = 44

44/6 = 2 R2
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Re: What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6  [#permalink]

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12 Nov 2016, 06:24
cyclicity of remainder is

8^1=2
8^2=4
8^3=2
8^4=4
all odd degrees of 8 has remainder 2 and all even has 4. So 8^15 has remainder 2

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Re: What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6  [#permalink]

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12 Nov 2016, 12:29
3
1
susheelh wrote:
What is the remainder when $$8^1+ 8^2+ 8^3……+8^{15}$$ is divided by 6

A. 0
B. 1
C. 2
D. 4
E. 5

$$\frac{8}{6}$$ = Remainder 2

$$\frac{8^2}{6}$$ = Remainder 4

$$\frac{8^3}{6}$$ = Remainder 2

$$\frac{8^4}{6}$$ = Remainder 4

Thus odd powers will have remainder 2 ; and even powers will have remainder 4

Now, $$8^1+ 8^2+ 8^3……+8^{15}$$ will have the following powers -

Odd = 1 , 3 , 5 , 7 , 9 , 11 , 13 , 15 ( 8 odd powers ) ; Sum of remainder = 16

Even Powers = 2, 4 , 6 , 8 , 10 , 12 , 14 ( 7 even Powers ) ; Sum of remainder = 28

Total sum of remainder = 44

44/6 = Remainder 2

Hence correct aswer will be (C) 2....

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Re: What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6  [#permalink]

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03 Dec 2016, 07:37
susheelh wrote:
What is the remainder when $$8^1+ 8^2+ 8^3……+8^{15}$$ is divided by 6

A. 0
B. 1
C. 2
D. 4
E. 5

$$8 = 2$$ (mod 6) ---> “leaves remainder 2 when divided by 6”.

In mod 6 our expression equals to:

$$\frac{2^1 + 2^2 + 2^3 + … + 2^{15}}{6} = \frac{2*(2^{15}-1)}{(2 – 1)*6} = \frac{2*(2^{15} – 1)}{2*3} = \frac{2^{15} – 1}{3}$$

Now $$2 = -1$$ (mod 3) and we have

$$\frac{(-1)^{15} – 1}{3} = \frac{-1 – 1}{3} = \frac{-2}{3}$$

Remainder -2 is the same as remainder 1 when divided by 3 but we need to multiply by cancelled factor 2

$$\frac{1}{3} = \frac{2}{6}$$

Our remainder is $$2$$.

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What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6  [#permalink]

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04 Dec 2016, 00:34
Here is my Take on this Question=>
USE PATTERN RECOGNITION

In Questions such as the one above,observing the pattern can be really helpful/
8=> k=6k+2
8+64=>6k
8+64+512=> 6k+2
8+64+512+4096=>6k
Hence for the number of terms being odd => Remainder is 2 and for the number of terms being even => Remainder is 0.
Since 15=odd
The remainder must be 2

Hence C

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Re: What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6  [#permalink]

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05 Aug 2018, 17:58
susheelh wrote:
What is the remainder when $$8^1+ 8^2+ 8^3……+8^{15}$$ is divided by 6

A. 0
B. 1
C. 2
D. 4
E. 5

$$8^1+ 8^2+ 8^3……+8^{15}$$ is same as
$$2^3+ 2^6+ 2^9……+2^{45}$$
What is the remainder when expression above is divided by 6?

To simplify, what is the remainder of
$$2^2+ 2^5+ 2^8……+2^{44}$$
when divided by 3? (Once we find the remainder, remember to multiply it by 2. Since we're factoring out 2 to make the expression easier to work with, we have to re-insert the factor of 2 at the end.)

Pattern:
1st term: r=1
2nd term: r=2
3rd term: r=1
etc
This means that the sum of every 2 terms will have r=0 (when div by 3). So, if there is an even number of terms, r=0; if there is an odd number of terms, r=1.

Total # of terms: $$\frac{44-2}{3}+1=15$$
So r=1 when div by 3. Re-introduce the 2 we factored out earlier, and r=2 when div by 6.

Re: What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6 &nbs [#permalink] 05 Aug 2018, 17:58
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