Remainder Question??

Note that each of 9^1, 9^2,...9^8 is an odd multiple of 3. The sum of 8 such odd multiples of 3 is an even multiple of 3, i.e. a multiple of 6. When a multiple of 6 is divided by 6, the remainder is 0

I also arrived the same answer as kevin using geometric progression..

Sum = a *(r^n) - 1/ (r-1)
n=8, r=9.

Sum = 9* (9^8 -1) / (9-1). Since 9 is odd, 9^8 is also odd. But 9^8-1 becomes even,hence can be represented as say 2k.
Sum = 3*3*2k/ 8. Clearly this is divisible by 3 and 2. Hence remainder =0.

@ACE_GRE

Sum = a *(r^n) - 1/ (r-1)
n=8, r=9.

Sum = 9* (9^8 -1) / (9-1). Since 9 is odd, 9^8 is also odd. But 9^8-1 becomes even,hence can be represented as say 2k.
Sum = 3*3*2k/ 8. Clearly this is divisible by 3 and 2. Hence remainder =0.

No Offence meant:
Just a thought to your statement Sum = 3*3*2k/ 8. Clearly this is divisible by 3 and 2. Hence remainder =0.

Is 3*3*2K/8 divisible by 6. where K is an Odd integer?
IMO
Obviously Yes and No both.
Depending on the value of K.
If K is a multiple of 8 yes it is divisible by 6
If K is not a multiple of 8 there will be a remainder other than 0

Thus we cannot use this logic. Either we wil have to solve the complete thing if we use the Sum of Gp or have to break the above to have the factors of 8 as well.

"No Offence meant: " None taken..

Thanks for pointing it out. I rushed at the answer. Using GP will take some long computation!

kevincanspain wrote:Note that each of 9^1, 9^2,...9^8 is an odd multiple of 3. The sum of 8 such odd multiples of 3 is an even multiple of 3, i.e. a multiple of 6. When a multiple of 6 is divided by 6, the remainder is 0

Could you please elaborate on the red part

I believe this is what he means.

the ones digit for 9^1, 9^2 ... 9^8, follows a pattern of 1, 9 , 1, 9 ... . if you add up an even amount (8 in this case) of odd digits, you will get an even digit.

an even digit that is divisible by 3 should also be divisible by six, because it will have a factor of 2 and 3.

apoorva.srivastva wrote:

kevincanspain wrote:Note that each of 9^1, 9^2,...9^8 is an odd multiple of 3. The sum of 8 such odd multiples of 3 is an even multiple of 3, i.e. a multiple of 6. When a multiple of 6 is divided by 6, the remainder is 0

Could you please elaborate on the red part

Odd + Odd = even

Odd + Odd + Odd + Odd + Odd + Odd + Odd + Odd = Even


For a number to be a multiple of 6 - It has to be even and it has to be a multiple of three

Hope that helps

9^1 + 9^2 + 9^3 + ... + 9^8 can be represented as
(3²+ 3^4+........3^16) which is to be divided by 6 or (3x2)
(3² + 3^4+........3^16) / 3x2
the expression (3², 3^4+........3^16) is definitely divided by 3
and since the expression has even no of odd terms i.e 8 odd terms it is divisible by 2 as explained by kevincanspain
so the remainder will be 0.