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

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Director
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What is the remainder when 7^345 +7^11 -2 is divided by 7 [#permalink]  20 Feb 2006, 22:30
What is the remainder when
7^345 +7^11 -2 is divided by 7?
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Joined: 07 Jul 2004
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(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
Director
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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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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.
Director
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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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Success is my only option, failure is not -- Eminem

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Joined: 07 Jul 2004
Posts: 5078
Location: Singapore
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Kudos [?]: 184 [0], given: 0

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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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
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