Remainder Problem

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Remainder Problem

by Aman verma » Wed Feb 17, 2010 3:26 am
Q: The remainder when 888222888222888222........... upto 9235 digits is divided by 5^3 ( 5 raised to the power 3) is :

a) 38

b) 72

c) 103

d) 139

e) Can't be determined

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by ajith » Wed Feb 17, 2010 3:55 am
Aman verma wrote:Q: The remainder when 888222888222888222........... upto 9235 digits is divided by 5^3 ( 5 raised to the power 3) is :

a) 38

b) 72

c) 103

d) 139

e) Can't be determined
888222888222888222........... upto 9235

The last 7 digits of this number - 8882228

125 is a factor of 1000

888222888222888222........... upto 9232 digits followed by 3 zeros are divisible by 125

now the question is what is the remainder when 228 is divided by 125

it is 103
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by kvamsy » Wed Feb 17, 2010 3:56 am
Aman verma wrote:Q: The remainder when 888222888222888222........... upto 9235 digits is divided by 5^3 ( 5 raised to the power 3) is :

a) 38

b) 72

c) 103

d) 139

e) Can't be determined
I am getting as OA : A, Please let me know if this is correct to post the explanation

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by Aman verma » Wed Feb 17, 2010 10:53 am
ajith wrote:
Aman verma wrote:Q: The remainder when 888222888222888222........... upto 9235 digits is divided by 5^3 ( 5 raised to the power 3) is :

a) 38

b) 72

c) 103

d) 139

e) Can't be determined
888222888222888222........... upto 9235

The last 7 digits of this number - 8882228

125 is a factor of 1000

888222888222888222........... upto 9232 digits followed by 3 zeros are divisible by 125

now the question is what is the remainder when 228 is divided by 125

it is 103
But how did you arrive at the last 7 digits of this number - 8882228 .And why the last 7 digits ,why not 6 or 8 digits ?

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by ajith » Wed Feb 17, 2010 10:59 am
Aman verma wrote: But how did you arrive at the last 7 digits of this number - 8882228 .And why the last 7 digits ,why not 6 or 8 digits ?
the pattern is 888666 repeating i.e. a block of 6 repeating.
there are a total of 9235 digits

9235/6 leaves a remainder of 1 so, it ends 8882228

7 is 6+1, (6 of the block and one of the remainder)

now I am only interested in last 3 digits because 125 divides 1000 evenly.

Last 3 digits are 2 2 and 8 and finding out last 7 digits was a part of the plan to find last 3 digits. If you have a better method to find last 3 digits, It is fine.

(I do not care last 3,4,5,6,7,8,9 digits for that matter, all I am interested is last 3 digits)
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by Aman verma » Wed Feb 17, 2010 11:06 am
ajith wrote:
Aman verma wrote: But how did you arrive at the last 7 digits of this number - 8882228 .And why the last 7 digits ,why not 6 or 8 digits ?
the pattern is 888666 repeating i.e. a block of 6 repeating.
there are a total of 9235 digits

9235/6 leaves a remainder of 1 so, it ends 8882228

7 is 6+1, (6 of the block and one of the remainder)

now I am only interested in last 3 digits because 125 divides 1000 evenly.

Last 3 digits are 2 2 and 8 and finding out last 7 digits was a part of the plan to find last 3 digits. If you have a better method to find last 3 digits, It is fine.

(I do not care last 3,4,5,6,7,8,9 digits for that matter, all I am interested is last 3 digits)


You are the real champ !! I don't think anybody could provide a better explanation . I didn't had the explanation, only had the net answer. C it is.