Find the sum of the last three digits of 9 1 0 5 .
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Same method ;-)
Do 9 1 0 5 m o d 1 0 0 0
We know that 9 5 ≡ 4 9 ( m o d 1 0 0 0 )
So 9 1 0 5 = ( 9 5 ) 2 1 ≡ 4 9 2 1 ( m o d 1 0 0 0 )
We also know that 4 9 3 m o d 1 0 0 0 = 6 4 9
So 4 9 2 1 = ( 4 9 3 ) 7 ≡ 6 4 9 7 ( m o d 1 0 0 0 )
Another thing we know: 6 4 9 2 m o d 1 0 0 0 = 2 0 1
So 6 4 9 7 = ( 6 4 9 2 ) 3 ⋅ 6 4 9 ≡ 2 0 1 3 ⋅ 6 4 9 ( m o d 1 0 0 0 )
And so 6 4 9 7 ≡ 6 0 1 ⋅ 6 4 9 ( m o d 1 0 0 0 )
≡ 3 9 0 0 4 9 ( m o d 1 0 0 0 )
≡ 4 9 ( m o d 1 0 0 0 ) by the law that A ≡ B ( m o d n ) is equal to A m o d n = B m o d n
The last 3 digits are 0 4 9 , so the sum of the digits are 1 3
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We write 9 1 0 5 = ( 1 0 − 1 ) 1 0 5 .
We then apply the binomial theorem. Terms that have powers of 10 that are greater than 1 0 2 do not contribute to the last three digits.
However, ( 2 1 0 5 ) 1 0 2 = 5 4 6 0 0 0 and doesn't contribute to the last 3 digits either.
Therefore, the only terms that contribute to the last 3 digits is 1 0 5 ⋅ 1 0 − 1 = 1 0 4 9 .
Therefore, the last 3 digits are 049, and our answer is 0 + 4 + 9 = 1 3 .