What is the remainder when

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guerrero: Master | Next Rank: 500 Posts; Posts: 227; Joined: Sun Apr 08, 2012 4:53 amFollowed by: 3 members; Thanked: 12 times

by guerrero » Mon Aug 05, 2013 10:37 pm

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

A. 0
B. 3
C. 2
D. 5
E. None of the above

OA B

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Source: Beat The GMAT — Problem Solving | Mon Aug 05, 2013 10:37 pm

macattack: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Wed Jun 26, 2013 10:55 pmFollowed by: 1 member; Thanked: 11 times

by macattack » Mon Aug 05, 2013 11:11 pm

Dear guerrero,

You need to know the rule of remainders which states that the remainder of (a+b):c= R(a:c)+R(b:c) keeping in mind that you should adjust it every time R gets bigger than c you subtract c and you get your remainder.
Here is an illustration: R(11:4)=3 and R(10:4)=2---> R((10+11):4)=5 but 5 is greater than 4 so subtract 4 and you get R=1. If you check the answer: 11+10=21---> 21:4 has indeed a remainder of 1.

Here is how you should attack such a question:

Remainder of 9^1:6=3
Remainder of 9^2:6=R 81:6=3
Remainder of 9^3:6= R 729:6=3

Notice the pattern? Now apply the remainder rule:
R((9^1+9^2+9^3+9^4+9^5+9^6+9^7+9^8+9^9):6)=R((9^1):6)+R((9^2):6)...R((9^9):6)=
3+3+3+3+3+3+3+3+3=27
27 is greater than 6 so adjust by subtracting 6 untill R < 6
27-6=21
21-6=15
15-6=9
9-6=3 (3 is less than six) so this is the remainder
The correct answer is hence B

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Tue Aug 06, 2013 12:41 am

all these "rules"... geez. if i had a memory like that, i'd rule the world.

how about this instead:

9^1, 9^2, etc. are all multiples of 3.
they're also all odd.
there are nine of them.
so...
when you add them together, you get another multiple of 3 that is odd.

the even multiples of 3 are ... the multiples of 6.
so, this sum is 3 more than a multiple of 6.
so the remainder is 3.

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macattack: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Wed Jun 26, 2013 10:55 pmFollowed by: 1 member; Thanked: 11 times

by macattack » Tue Aug 06, 2013 12:47 am

Wait a second...the sum of multiples of 3 is in fact a multiple of 3 and wait another second the sum of 9 odd numbers is an odd number since odd+odd=even and odd+even=odd 2 rules that you should know. lunarpower I agree rules are not fun but they will harass us as long as were aiming to score high on quant because they are indispensable in math!

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Tue Aug 06, 2013 1:04 am

macattack wrote:Wait a second...the sum of multiples of 3 is in fact a multiple of 3 and wait another second the sum of 9 odd numbers is an odd number since odd+odd=even and odd+even=odd 2 rules that you should know. lunarpower I agree rules are not fun but they will harass us as long as were aiming to score high on quant because they are indispensable in math!

i'll grant you that "odd + odd = odd", etc., are rules that you need to know.

as for the multiples of 3, i don't memorize that sort of thing (because i wouldn't retain it anyway). instead, i just think about literal groups of three things packaged together -- if i add up a bunch of those, then i still have a bunch of groups of three things.

same thing with other "rules".
i'm probably biased here, because i have a really bad memory, but i just don't like to try to remember stuff -- unless it can't be derived on the fly (e.g., the pythagorean theorem -- ok, even i have to memorize that one).

but in the case of the remainder rule stated above, i don't see any utility in remembering that separately as a rule, because i can just think about it: remainders are "leftovers". if i'm making packs of N things, then i can just add the leftovers, until i have N of them (at which point i can just make another pack of N).

in one sense, you and i are talking about the same thing here, since ultimately we're talking about the same facts.
but, in another sense, we're talking about very different things.
see, here's where my way of thinking wins: my mind isn't cluttered by a bunch of memorization, so i'm more likely to be able to discover other rules/patterns, via the same spirit of investigation.

in any case, the name of the game is "accumulate as many methods as possible", so, anyone reading this thread should try to understand (and produce) both my solution and the other poster's.

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mgm: Senior | Next Rank: 100 Posts; Posts: 46; Joined: Sun Sep 12, 2010 12:32 pmFollowed by: 1 member; Thanked: 1 times

by mgm » Tue Aug 06, 2013 4:02 am

I seem to be getting a different answer

Restating :

9(1 + 9 + 9^2 + 9^3 ...+9^8) / 6

==> 1 +9 units digit 0
9^2 + 9^3 units digit 0 and so on...

We are left with a units digit of 9^8 which is 1 ... that times 3 / 2 and with a remainder of 1 .. not sure where I went wrong...

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Tue Aug 06, 2013 4:56 am

mgm wrote:I seem to be getting a different answer

Restating :

9(1 + 9 + 9^2 + 9^3 ...+9^8) / 6

==> 1 +9 units digit 0
9^2 + 9^3 units digit 0 and so on...

We are left with a units digit of 9^8 which is 1 ... that times 3 / 2 and with a remainder of 1 .. not sure where I went wrong...

oh, no, you can't do that with remainders. i.e., you can't "reduce" a division ratio. (this is what it appears you're trying to do, anyway -- it looks as though you're trying to take the 9/6 and "reduce" it to 3/2.
nope. not with remainders.

think about small numbers and this will be clear:
it's true that 10/4 = 5/2. however...
* when you divide 10 by 4, the remainder is 2.
* when you divide 5 by 2, the remainder is 1.
oops.

(also, if you think conceptually about what a remainder is, this should make sense, too. e.g., if i distribute 10 items into packs of 4, then i'm going to have 2 items left over.
if i distribute 5 items into packs of 2, then that's a completely different situation, in which i have 1 item left over.)

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faraz_jeddah: Master | Next Rank: 500 Posts; Posts: 358; Joined: Thu Apr 18, 2013 9:46 amFollowed by: 7 members; Location: Jeddah, Saudi Arabia; Thanked: 42 times; GMAT Score:730

by faraz_jeddah » Tue Aug 06, 2013 6:32 am

lunarpower wrote:
9^1, 9^2, etc. are all multiples of 3.
they're also all odd.
there are nine of them.
so...
when you add them together, you get another multiple of 3 that is odd.

Why did you take 3? Should we not say multiples of 9?

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Aug 06, 2013 8:40 am

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

A. 0
B. 3
C. 2
D. 5
E. None of the above

OA B

9Â¹ + 9Â² + 9Â³ + 9â�´ + 9â�µ + 9â�¶ + 9â�· + 9â�¸ + 9â�¹ = the sum of 9 multiples of 9 = an ODD MULTIPLE OF 9.

Divide odd multiples of 9 by 6 and LOOK FOR A PATTERN.
9/6 = 1 R3.
27/6 = 4 R3.
45/6 = 7 R3.
63/6 = 10 R3.
In each case, the remainder is 3.

The correct answer is B.

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Wed Aug 07, 2013 6:48 am

faraz_jeddah wrote:
lunarpower wrote:
9^1, 9^2, etc. are all multiples of 3.
they're also all odd.
there are nine of them.
so...
when you add them together, you get another multiple of 3 that is odd.

:shock:

Why did you take 3? Should we not say multiples of 9?

you can actually do either.
again, this is a great example of "just try whatever comes to mind" -- one of the awesome things about GMAT problems is that they can often be solved in many different ways.

the honest reason why i thought in terms of 3 here is "it's just what popped into my head".
if you pressed me for a justification, i'd say something like "it's a factor that 6 and 9 have in common", but, really, i just started playing with the first number that came to mind.

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shailendra.sharma: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Tue Feb 26, 2013 3:49 am; Thanked: 3 times

by shailendra.sharma » Thu Aug 08, 2013 8:43 am

I did in a bit different way, but more or less similar.

6 = 2 x 3

9Â¹ + 9Â² + 9Â³ + 9â�´ + 9â�µ + 9â�¶ + 9â�· + 9â�¸ + 9â�¹ are all fully divisible by 3
==> and what we get is multiples of 3 or simply 9 odd numbers
==> any odd divided by 2, gives remainder of 1
==> you have nine 1s as remainder
==> add them together to get 9
==> adjust remainder to get final remainder as 3

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by Matt@VeritasPrep » Thu Aug 08, 2013 12:01 pm

Shailendra, I think that's a pretty clever solution, and one that would work well applied to other numbers too.

The easiest way to do this one, at least for me, is to consider it like this:

Any odd multiple of 9 can be written as 9 * (2k+1), where k is some integer.

9 * (2k+1) = 18k + 9

18k + 9 = 18k + 6 + 3

(18k + 6) will be divisible by 6, as both 18k and 6 divide by 6, so the remainder is the final +3.

Piece of cake!

This method can also show why any even multiple of 9 is divisible by 6.

An even multiple of 9 = 9 * (2k), where k is some integer.

18k / 6 = 3k, or an integer, so any even multiple of 9 is divisible by 6.

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Fri Aug 09, 2013 12:39 am

shailendra.sharma wrote:I did in a bit different way, but more or less similar.

6 = 2 x 3

9Â¹ + 9Â² + 9Â³ + 9â�´ + 9â�µ + 9â�¶ + 9â�· + 9â�¸ + 9â�¹ are all fully divisible by 3
==> and what we get is multiples of 3 or simply 9 odd numbers
==> any odd divided by 2, gives remainder of 1
==> you have nine 1s as remainder
==> add them together to get 9
==> adjust remainder to get final remainder as 3

so, the red part here -- if you actually mean division by 2 -- is an issue. why are you considering remainders on division by 2?
you can't do that; the division in the problem is division by 6, not by 2.

as an illustration of what could go wrong here, let's say it's
3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 5
instead of what's there right now.
if you're saying what i think you're saying, then you'd attempt to make the same argument, and you would conclude that the remainder (when that sum is divided by 6) would be 3 again.
this time, however, you'd be incorrect; the remainder is 5 this time. (if you don't see this, just add up the numbers -- the sum is 29, and 29/6 has a remainder of 5.)

in general -- if a problem involves remainders on division by "n", then be VERY cautious about dividing by any number other than "n". as you can see here, that kind of reasoning often doesn't work.

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shailendra.sharma: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Tue Feb 26, 2013 3:49 am; Thanked: 3 times

by shailendra.sharma » Fri Aug 09, 2013 6:04 am

lunarpower wrote:
shailendra.sharma wrote:I did in a bit different way, but more or less similar.

6 = 2 x 3

9Â¹ + 9Â² + 9Â³ + 9â�´ + 9â�µ + 9â�¶ + 9â�· + 9â�¸ + 9â�¹ are all fully divisible by 3
==> and what we get is multiples of 3 or simply 9 odd numbers
==> any odd divided by 2, gives remainder of 1
==> you have nine 1s as remainder
==> add them together to get 9
==> adjust remainder to get final remainder as 3
so, the red part here -- if you actually mean division by 2 -- is an issue. why are you considering remainders on division by 2?
you can't do that; the division in the problem is division by 6, not by 2.

as an illustration of what could go wrong here, let's say it's
3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 5
instead of what's there right now.
if you're saying what i think you're saying, then you'd attempt to make the same argument, and you would conclude that the remainder (when that sum is divided by 6) would be 3 again.
this time, however, you'd be incorrect; the remainder is 5 this time. (if you don't see this, just add up the numbers -- the sum is 29, and 29/6 has a remainder of 5.)

in general -- if a problem involves remainders on division by "n", then be VERY cautious about dividing by any number other than "n". as you can see here, that kind of reasoning often doesn't work.

You are right... it's good I posted my answer and have got the correction in my thinking

Thanks !

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