Algebra Question #15: Application of Remainder Theorem
Let R be the remainder of 3 1 6 1 0 1 − 1 Find R
The answer is 0.
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5 solutions
∵ 1 6 = 1 m o d 3 ∴ 1 6 1 0 1 − 1 = 1 1 0 1 − 1 m o d 3 = 0 m o d 3
( 1 6 1 0 1 − 1 ) / 3 = ( 4 2 0 2 − 1 ) / ( 4 − 1 ) Let x = 4 => ( x 2 0 2 − 1 ) / ( x − 1 ) By Remainder Theorem, R = 0
just a suggestion write,
4 − 1 4 2 0 2 − 1 as 1 + 4 + 4 2 + . . . . + 4 2 0 1
So its easy to understand if one do'nt know
Can you please explain the remainder theorem part in a bit more detail. I forgot this method >_<
try either 1 ,2 0r 0 as there are 3 tries
we break 16 into (15+1) So the numerator becomes (15+1)^101 - 1^101 = 15^101 Therefore any power of 15 when divided by 3 will always give remainder as zero
1 6 1 0 1 ≡ ( 1 5 + 1 ) 1 0 1 ≡ 1 ( m o d 3 )
∴ 1 6 1 0 1 − 1 ≡ 0 ( m o d 3 )