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