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

Question

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

GMATinsight · Accepted Answer

8^1+ 8^2+ 8^3 + ... +8^{15}  is divided by 6

Remainder when  8^1  is divided by 6,  R  = 2 
Remainder when  8^2  is divided by 6,  R  = 2^2 = 4 
Remainder when  8^3  is divided by 6,  R  = R  = 2 
Remainder when  8^4  is divided by 6,  R  = R  = 4 
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i.e. Series of remainders is  (2+4)+(2+4)+(2+4)+(2+4)+(2+4)+(2+4)+(2+4) + 2

But each of the  7 pairs (each in a bracket)  i.e. 14 terms in pairs sums up to 6 i.e., leaves remainder 0 when divided by 6

Hence, Final Remainder =  2

Answer: Option C

chetan2u · Answer

Something that came to me immediately, and should be true for terms with base one less than multiple of 3.
 If we take out 8 from two terms with consecutive powers, the number should be multiple of 6.  =>  8^a+8^{a+1}=8^a(1+8)=9*8^a , where a is positive integer.

8^1+ 8^2+ 8^3 + ... 8^{14}+8^{15}=8^1+8^2(1+8)+8^4(1+8)...8^14(1+8) 
All terms except 8^1 are divisible by 6. Thus, the remainder is 2.

C

PyjamaScientist · Answer

We need to check divisibility of  8^1+ 8^2+ 8^3 + ... +8^{15}  by 6.
 6 = 2 * 3 . Since, we know 8 is an even no. and thus it will no remainder when divided by 2, we need to basically check for divisibility of  8^1+ 8^2+ 8^3 + ... +8^{15}  by just no.  3 .

Using Binomial theorem, 
We can write  8^1+ 8^2+ 8^3 + ... +8^{15}  as =  (9-1)^1+ (9-1)^2+ (9-1)^3 + ... +(9-1)^{15} 
 (9-1)^k  for k odd, this will give remainder of -1 when divided by 3. And for k even,  (9-1)^k  will give remainder of +1 when divided by 3.

Thus, our series of remainder from individual  (9-1)^1+ (9-1)^2+ (9-1)^3 + ... +(9-1)^{15}  will be =  -1 + 1 - 1 + 1 ... +1 - 1 . Since, there are 15 (odd no.) terms, first 14 terms will cancel out, and the last term remaining would be  -1 .

Thus, the remainder would be,  -1 = 3(-1) +   2 . Option (C).

BrushMyQuant · Answer

We need to find the remainder when  8^1+ 8^2+ 8^3 + ... +8^{15}  is divided by 6?

To solve this problem we need to find the cycle of remainder of power of 8 when divided by 6

Remainder of  8^1  (=8) by 6 = 2
Remainder of  8^2  (=64) by 6 = 4
Remainder of  8^3  (=512) by 6 = 2
   Remainder of  8^4  (=4096) by 6 = 4

Now, notice that Cycle is 2 and Sum of two consecutive remainders = 2 + 4 = 6 which is same as remainder of 6 by 6 = 0

=> Remainder of  8^1+ 8^2+ 8^3 + ... +8^{14} +8^{15}  by 6 = Remainder of  8^1+ 8^2+ 8^3 + ... +8^{14} by 6 + Remainder of 8^{15}  by 6 = 0 + 2 = 2

So,  Answer will be C 
Hope it Helps!

Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem

https://www.youtube.com/watch?v=h1HTEhmUb1w

bumpbot · Answer

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

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