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What is the reminder when 91 + 92 + 93 + ...... + 99 is

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VP
Joined: 29 Apr 2003
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What is the reminder when 91 + 92 + 93 + ...... + 99 is [#permalink]  05 Apr 2006, 01:11
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100% (01:03) correct 0% (00:00) wrong based on 2 sessions
What is the reminder when 91 + 92 + 93 + ...... + 99 is divided by 6?

(1) 0
(2) 3
(3) 4
(4) None of these

Director
Joined: 05 Feb 2006
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The number is 95*9=855 divide by 6... 3 is the remainder...
VP
Joined: 29 Apr 2003
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Sorry I posted wrongly! My Mistake!! This is the question:

What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6?
Manager
Joined: 24 Oct 2005
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the remainder is 3....
I did the same procedure as SimaQ but using the exponents.
Does anyone know a shortcut for calculating this question?
VP
Joined: 29 Apr 2003
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Let me know when you guys are ready for the OA... I got the anwer wrong myself.. had to look up the solution!

SVP
Joined: 24 Sep 2005
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sm176811 wrote:
Sorry I posted wrongly! My Mistake!! This is the question:

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

There're some ways to solve this problem:
(1) notice that 9^n divided by 6 always has remainder of 3 ---> from 9^1 to 9^9, we have 9 terms --> the sum of all remainders = 3*9= 27 , 27 divided by 6 has remainder of 3 --> 3 is the answer

(2) notice that 6= 2*3
We have all terms of the sum are divisible by 3 --> the sum is divisible by 3
+ Because there're 9 terms and each term is odd --> the sum is odd ---> the remainder of the sum when it's divided by 6 can be 1,3 or 5
+ If the remainder of the sum divided by 6 is 1 or 5, the whole sum can't be divisible by 3
---> 3 is the only answer.
VP
Joined: 29 Apr 2003
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Well Option 2 (which is 3) is the answer!

This is the official expln:

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'.
Manager
Joined: 24 Oct 2005
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laxieqv...that was a great explanation.

Thanks,
VP
Joined: 28 Mar 2006
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9^1 + 9^2 + 9^3 + ...... + 9^9

can be rewritten as

9 + 9[9 + 9^2]+9^3[9 + 9^2].....

you get

9 + [(9^1 + 9^2)*a very large number which we dont care ]

Since 9+81 is 90 which is divisible by 6

so ultimately u are diving 9/6 whose remainder = 3
VP
Joined: 28 Mar 2006
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sm176811 wrote:
I like ur soln!

Thanks!...For series problems always search for a pattern
Senior Manager
Joined: 24 Jan 2006
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90+91+92+93+94+95+96+97+98+99
9(90)+(1+2+3+4+5+6+7+8+9)
9(90)+5(9)
9(95)/6
3(95)/2
==>Q=3
Senior Manager
Joined: 11 Nov 2005
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3.

each number (9 9^2, 9^3...) divided by 6 gives remainder 3,

3*9 =27

27/6, remainder = 3
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