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

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What is the reminder when 91 + 92 + 93 + ...... + 99 is [#permalink] New post 05 Apr 2006, 01:11
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What is the reminder when 91 + 92 + 93 + ...... + 99 is divided by 6?

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




Correct Answer - (2)
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 [#permalink] New post 05 Apr 2006, 01:18
Answer is 3....

The number is 95*9=855 divide by 6... 3 is the remainder...
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 [#permalink] New post 05 Apr 2006, 01:36
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?
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 [#permalink] New post 05 Apr 2006, 02:33
the remainder is 3....
I did the same procedure as SimaQ but using the exponents.
Does anyone know a shortcut for calculating this question?
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 [#permalink] New post 05 Apr 2006, 02:44
Let me know when you guys are ready for the OA... I got the anwer wrong myself.. had to look up the solution!






P.S.: The official answer is encrypted in this page!! Go an figure out what it is! ;)
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 [#permalink] New post 05 Apr 2006, 02:56
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.
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 [#permalink] New post 05 Apr 2006, 03:04
Well Option 2 (which is 3) is the answer!

:)


This is the official expln:

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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 [#permalink] New post 05 Apr 2006, 09:02
laxieqv...that was a great explanation.

Thanks,
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 [#permalink] New post 05 Apr 2006, 15:18
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
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 [#permalink] New post 05 Apr 2006, 15:23
sm176811 wrote:
I like ur soln!


Thanks!...For series problems always search for a pattern
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 [#permalink] New post 06 Apr 2006, 07:03
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
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 [#permalink] New post 06 Apr 2006, 12:15
3.

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

3*9 =27


27/6, remainder = 3
  [#permalink] 06 Apr 2006, 12:15
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