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What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is

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What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is [#permalink] New post 12 Nov 2005, 00:09
What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6

(a) 0
(b) 3
(c) 4
(d) 5
(e) 6
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 [#permalink] New post 12 Nov 2005, 01:54
9^1 + 9^2 + 9^3 + ...... + 9^9 is a geometric progression

and sum = 9 * ( 9^9 -1)/ 8

now 9^9 -1 /8 will always be an even integer.
so the result we get is 9* even integer

the result will be divisible by 6

so reminder will be 0.

A is the answer.
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Re: reminder [#permalink] New post 12 Nov 2005, 02:01
cool_jonny009 wrote:
What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6

(a) 0
(b) 3
(c) 4
(d) 5
(e) 6


9^n divided by 6 always has remainder of 3 ( i'll try to prove this)
from 9^1 to 9^9 we have 9 such numbers , thus the remainer added up to 27 or 3 ( 27= 6*4+3)
Thus, B it is.
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 [#permalink] New post 12 Nov 2005, 02:10
we have 9^n-3 = 3^(2n) -3 = 3* ( 3^(2n-1)-1) ( n is a positive integer)
3^(2n-1) is an odd number --> 3^(2n-1) -1 is an even number --->
3^(2n-1)-1 is divisible by 2.
----> 3* ( 3^(2n-1)-1) is divisible by 3*2 ( or 6) ----> 9^n-3 is divisible by 6 ----> 9^n divided by 6 has remainder of 3.
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 [#permalink] New post 12 Nov 2005, 02:17
laxieqv wrote:
we have 9^n-3 = 3^(2n) -3 = 3* ( 3^(2n-1)-1) ( n is a positive integer)
3^(2n-1) is an odd number --> 3^(2n-1) -1 is an even number --->
3^(2n-1)-1 is divisible by 2.
----> 3* ( 3^(2n-1)-1) is divisible by 3*2 ( or 6) ----> 9^n-3 is divisible by 6 ----> 9^n divided by 6 has remainder of 3.


yups you are right. answer should be B.

I just figured now 9^x -1 /8 will always be an even integer. when x is even

for example 9^2 -1/8 = 10

but when x is odd, 9^x -1/8 will be odd

so 9 * odd / 6 will always have a reminder of 3.
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 [#permalink] New post 12 Nov 2005, 08:18
This method took 25 seconds...

Add the total exponents: 9+8+7+6...1---->45

Look for a pattern: 9^1=9, 9^2=81, 9^3=729

So, any even exponent of 9 will yield 1 as the units digit.
And any odd exponent of 9 will yield 9 as the units digit.

45 is odd so the units digit must be 9.

9/6=1R3

Choose B
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 [#permalink] New post 12 Nov 2005, 08:29
Excellent shortcut! thanks
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 [#permalink] New post 12 Nov 2005, 09:31
Here is a shortcut that took 5 seconds....


Factor out the 9, we get

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

The only portion that contributes to the remainder is

9/6 -> with R = 3.

The answer is B.
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 [#permalink] New post 12 Nov 2005, 09:39
rigger wrote:
Here is a shortcut that took 5 seconds....


Factor out the 9, we get

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

The only portion that contributes to the remainder is

9/6
-> with R = 3.

The answer is B.


can you explain why? (the bold part)
I think it holds only when the part in ( ) is proved to be divisible by 6.
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 [#permalink] New post 12 Nov 2005, 10:11
Allright guys...
remainder when 9^1 + 9^2 + 9^3 + .... + 9^8 is divided by 6

the remainder is = 0

but when 9^1 + 9^2 + 9^3 + .... + 9^9 ( which was the orignal question )

remainder is = 3 ....


:done
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Re: reminder [#permalink] New post 13 Nov 2005, 20:10
cool_jonny009 wrote:
What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6

(a) 0
(b) 3
(c) 4
(d) 5
(e) 6


Hmmm this actually took me more than 2 minutes to think of a good way to solve.

9^1+..9^9=9*(1+..+9^8)
We know that 9 can be divided by 3, but not 2. So we need to know if (1+9+9^2+..9^8) can be divided by 2. We also know that 9^n would be odd, for all n>0. So all we need to do is to count how many odd number we have in there. We have 9^1 to 9^8, eight items, plus 1. So the sum of 9 odd numbers would be odd. Let say it is equal to 2m+1.
Than we can right the original expression like this:
9^1+9^2+...+9^9=9*(1+9^1+..9^8)=9*(2m+1)=12m+9
Now it's clear that the the reminder it divided by 6 is 3.
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 [#permalink] New post 14 Nov 2005, 08:18
when two numbers are div. by a divisor and its remainders are r1 and r2, the remainder of the sum of the two numbers is r1+r2=r3. when r1+r2>d, the remainder is r1+r2-d=r3. in this case the remainder is always 3. the sum of the remainders is 27. we have to substract the divisor until it is less than 6. thats 3 !
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 [#permalink] New post 16 Nov 2005, 18:13
I have no idea what you are doing here....?

can you explain mathematically?

GMATT73 wrote:
This method took 25 seconds...

Add the total exponents: 9+8+7+6...1---->45

Look for a pattern: 9^1=9, 9^2=81, 9^3=729

So, any even exponent of 9 will yield 1 as the units digit.
And any odd exponent of 9 will yield 9 as the units digit.

45 is odd so the units digit must be 9.

9/6=1R3

Choose B
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 [#permalink] New post 16 Nov 2005, 18:19
I just did it this way...

9/6 remainder is 3...

81/6 remainder is 3....

729/6 remainder is 3...after this point I know all the remaining numbers will give remainder of 3...

so 3+...3=27/6=reamainder is 3...

took about 1 minute to realize this...another 15 seconds to put it together ...nice question...
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 [#permalink] New post 16 Nov 2005, 19:09
I like the idea of factoring out 9/6. There is no reason why you can't do this right?

9/6* (1 + 9 + 9^2 + 9^3 + 9^4 +9^5 + 9^6 + 9^7 + 9^8)

then you can cleary see that the remainder will be 3. A lot of the other explantions I did not understand. Can you further explain the "add the exponents and they add up to 45" rationale?

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official explanation [#permalink] New post 17 Nov 2005, 12:54
here is the official explanation
the bold part sums up what have said
hope it helps ...

Solution:

Any number that is divisible by ‘6' will be a number that is divisible by both ‘2' and ‘3'. i.e. the number should an even number and should be divisible by three (sum of its digits should add to a multiple of ‘3').

Any number that is divisible by ‘9' is also divisible by ‘3' but unless it is an even number it will not be divisibly by ‘6'.

In the above case, ‘9' is an odd number. Any power of ‘9' which is an odd number will be an odd number which is divisible by ‘9'.

Therefore, each of the terms 91, 92 etc are all divisibly by ‘9' and hence by ‘3' but are odd numbers.

Any multiple of ‘3' which is odd when divided by ‘6' will leave a reminder of ‘3'.

For example 27 is a multiple of ‘3' which is odd. 27/6 will leave a reminder of ‘3'. Or take 45 which again is a multiple of ‘3' which is odd. 45/6 will also leave a reminder of ‘3'.

Each of the individual terms of the given expression 91 + 92 + 93 + ...... + 99 when divided by 6 will leave a reminder of ‘3'. There are 9 such terms. The sum of all the reminders will therefore be equal to 9*3 = 27.
27/6 will leave a reminder of ‘3'.

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official explanation   [#permalink] 17 Nov 2005, 12:54
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