888222888222888222........... upto 9235Aman 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 explanationAman 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
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 ?ajith wrote:888222888222888222........... upto 9235Aman 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
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
the pattern is 888666 repeating i.e. a block of 6 repeating.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 ?
ajith wrote:the pattern is 888666 repeating i.e. a block of 6 repeating.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 ?
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)
