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What is the remainder when 7^345 +7^11 -2 is divided by 7

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What is the remainder when 7^345 +7^11 -2 is divided by 7 [#permalink] New post 20 Feb 2006, 23:30
What is the remainder when
7^345 +7^11 -2 is divided by 7?
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 [#permalink] New post 20 Feb 2006, 23:43
(7^345 + 7^12 - 2)/7 = 7^345 + 7^12/7 - 2/7

7^345 + 7^12/7 will result in a whole number

x.000000 - 2/7 --> Remainder will be approximately 1-2/7 =0.714
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 [#permalink] New post 20 Feb 2006, 23:50
ywilfred wrote:
(7^345 + 7^12 - 2)/7 = 7^345 + 7^12/7 - 2/7

7^345 + 7^12/7 will result in a whole number

x.000000 - 2/7 --> Remainder will be approximately 1-2/7 =0.714


I think I am gonna quote you again.. It will be 5.. think about it this way..

the first two are divisible by 7.. so remainder is gonna be 0 for themm.. right?? Now.. its like (49 +14 -2)/7 or (49 + 49 -2)/7.. No matter what remainder will always be 5..
ps: I checked it with scientific calc
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 [#permalink] New post 20 Feb 2006, 23:56
bewakoof wrote:
ywilfred wrote:
(7^345 + 7^12 - 2)/7 = 7^345 + 7^12/7 - 2/7

7^345 + 7^12/7 will result in a whole number

x.000000 - 2/7 --> Remainder will be approximately 1-2/7 =0.714


I think I am gonna quote you again.. It will be 5.. think about it this way..

the first two are divisible by 7.. so remainder is gonna be 0 for themm.. right?? Now.. its like (49 +14 -2)/7 or (49 + 49 -2)/7.. No matter what remainder will always be 5..
ps: I checked it with scientific calc


I thought of it this way:

The first two are always whole numbers, I think we're on the same page on that. So it will be like:

X.00000 - 2/7 (where X is the whole number and the sting of decimal digits behind). Since 2/7 wil not have a whole number in front, to subtract we need to carry '1' from X and the string of digits at the end will look like 1-2/7 = 0.71..... which was how I got my answer.
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 [#permalink] New post 21 Feb 2006, 00:03
ywilfred wrote:

I thought of it this way:

The first two are always whole numbers, I think we're on the same page on that. So it will be like:

X.00000 - 2/7 (where X is the whole number and the sting of decimal digits behind). Since 2/7 wil not have a whole number in front, to subtract we need to carry '1' from X and the string of digits at the end will look like 1-2/7 = 0.71..... which was how I got my answer.


remainder can't be 0.71.. Remainder will always be an Integer.. I hope we both agree on that.. the bolded portion does not hold true.. i think.. you can't carry over 1..like that..
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 [#permalink] New post 21 Feb 2006, 00:07
bewakoof wrote:
ywilfred wrote:

I thought of it this way:

The first two are always whole numbers, I think we're on the same page on that. So it will be like:

X.00000 - 2/7 (where X is the whole number and the sting of decimal digits behind). Since 2/7 wil not have a whole number in front, to subtract we need to carry '1' from X and the string of digits at the end will look like 1-2/7 = 0.71..... which was how I got my answer.


remainder can't be 0.71.. Remainder will always be an Integer.. I hope we both agree on that.. the bolded portion does not hold true.. i think.. you can't carry over 1..like that..


Didn't read the remainder part. But 'carry' over of 1 is legal. Say 4.0 - 0.3, we need to carry '1' from the '4' to the '0' so we have 3.(1)0 - 0.3 = 3.7. The last digit '7' is akin to having 1-0.3 = 0.7. But I agree my answer is valid only if it's not asking for a remainder.
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 [#permalink] New post 21 Feb 2006, 02:32
Agree with 5

I broke the stem down to

7^1 + 7^3 -2, which results in the same lasts digits ( the pattern goes 7,49,xx3, xxx1,xxx7)

7^3, has no remainder

7^1 - 2 leaves 5

I think this method is fast and secure
  [#permalink] 21 Feb 2006, 02:32
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