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3 solutions
9 9 9 9 1 0 0 0 = ( 1 0 0 0 0 − 1 ) 1 0 0 0
then by binomial expansion we can say that there are 1001 terms out of which 1000 are divisible by 5 except the last term which is ( − 1 ) 1 0 0 0 = 1which is our remainder
9
9
9
9
=
9
9
×
1
0
1
Using Congruency Theory,
9
9
=
(
−
1
)
(
m
o
d
5
)
&,
1
0
1
=
1
(
m
o
d
5
)
Therefore,
9
9
×
1
0
1
=
(
−
1
×
1
)
(
m
o
d
5
)
⇒
9
9
9
9
=
−
1
(
m
o
d
5
)
⇒
9
9
9
9
1
0
0
0
=
(
−
1
)
1
0
0
0
(
m
o
d
5
)
⇒
9
9
9
9
1
0
0
0
=
1
(
m
o
d
5
)
So, remainder would be 1.
The units digits of the powers of 9 have a pattern: 9^1 = 9, 9^2 = 81, 9^3 = 729, 9^4 = 6561, and so on.
For even powers of 9, the number ends in a 1.
All natural numbers that end in 1 have a remainder of 1 when divided by 5. So, our answer is 1