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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]

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12 Nov 2005, 01:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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12 Nov 2005, 02:54
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.

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12 Nov 2005, 03:01
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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12 Nov 2005, 03:10
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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12 Nov 2005, 03:17
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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12 Nov 2005, 09:18
This method took 25 seconds...

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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12 Nov 2005, 09:29
Excellent shortcut! thanks
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12 Nov 2005, 10:31
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.

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12 Nov 2005, 10:39
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.

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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12 Nov 2005, 11:11
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 ....

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13 Nov 2005, 21:10
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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14 Nov 2005, 09:18
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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16 Nov 2005, 19:13
I have no idea what you are doing here....?

can you explain mathematically?

GMATT73 wrote:
This method took 25 seconds...

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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16 Nov 2005, 19:19
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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16 Nov 2005, 20:09
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?

Thanks
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17 Nov 2005, 13:54
here is the official explanation
the bold part sums up what have said
hope it helps ...

Solution:

Any number that is divisible by â€˜6' will be a number that is divisible by both â€˜2' and â€˜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 â€˜3').

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

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

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

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

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

Each of the individual terms of the given expression 91 + 92 + 93 + ...... + 99 when divided by 6 will leave a reminder of â€˜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 â€˜3'.

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official explanation   [#permalink] 17 Nov 2005, 13:54
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# What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is

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