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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.

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.

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. _________________

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. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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 ! _________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

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?

official explanation [#permalink]
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 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'. _________________

when there is a will there is a way

best regards

gmatclubot

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

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