Remainder Question??

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apoorva.srivastva: Master | Next Rank: 500 Posts; Posts: 146; Joined: Wed Aug 27, 2008 5:41 am; Thanked: 3 times

Remainder Question??

by apoorva.srivastva » Thu Jan 07, 2010 12:52 pm

What is the remainder when 9^1 + 9^2 + 9^3 + .... + 9^8 is divided by 6?

(1) 3
(2) 2
(3) 0
(4) 5
(5.)none of these

please explain your reasoning

Source: Beat The GMAT — Problem Solving | Thu Jan 07, 2010 12:52 pm

kevincanspain: GMAT Instructor; Posts: 613; Joined: Thu Mar 22, 2007 6:17 amFollowed by: 64 members; Location: madrid; Thanked: 171 times; GMAT Score:790

by kevincanspain » Thu Jan 07, 2010 2:47 pm

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

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ace_gre: Senior | Next Rank: 100 Posts; Posts: 98; Joined: Mon Nov 23, 2009 2:30 pmFollowed by: 1 member; Thanked: 26 times

by ace_gre » Thu Jan 07, 2010 4:10 pm

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.

raviraushanjha: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Sat Jan 02, 2010 6:59 pm

by raviraushanjha » Thu Jan 07, 2010 4:36 pm

@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.

ace_gre: Senior | Next Rank: 100 Posts; Posts: 98; Joined: Mon Nov 23, 2009 2:30 pmFollowed by: 1 member; Thanked: 26 times

by ace_gre » Thu Jan 07, 2010 9:30 pm

"No Offence meant: " None taken..

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

apoorva.srivastva: Master | Next Rank: 500 Posts; Posts: 146; Joined: Wed Aug 27, 2008 5:41 am; Thanked: 3 times

by apoorva.srivastva » Thu Jan 07, 2010 9:59 pm

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

at325: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Mon Mar 29, 2010 9:16 pm

by at325 » Mon Mar 29, 2010 9:22 pm

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.

ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Mon Mar 29, 2010 11:27 pm

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

Always borrow money from a pessimist, he doesn't expect to be paid back.

kstv: Legendary Member; Posts: 610; Joined: Fri Jan 15, 2010 12:33 amFollowed by: 2 members; Thanked: 47 times

by kstv » Tue Mar 30, 2010 3:38 am

9^1 + 9^2 + 9^3 + ... + 9^8 can be represented as
(3Â²+ 3^4+........3^16) which is to be divided by 6 or (3x2)
(3Â² + 3^4+........3^16) / 3x2
the expression (3Â², 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.

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