Radical powers
Find the last three digits of
N = ( 1 6 1 + 1 6 1 2 − 1 ) 3 4 + ( 1 6 1 + 1 6 1 2 − 1 ) − 3 4 .
The answer is 207.
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1 solution
This was not my intended solution; it is possible to determine the exact value without resorting to approximation.
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Got it, we let M = ( 1 6 1 + 1 6 1 2 − 1 ) 3 1 + ( 1 6 1 − 1 6 1 2 − 1 ) 3 1
Cube both sides and simplify like what did above for N , we get a cubic equation, we show that the equation have linear factor of ( M − 7 ) and another quadratic factor with a negative discriminant, so M = 7 only, with this, we get a relation of N = ( M 2 − 2 ) 2 − 2 = 2 2 0 7
I also came upon this cubic and got the answer the same way...is there a better method to do this?
Because 1 6 1 2 − ( 1 6 1 2 − 1 ) = 1 , then ( 1 6 1 + 1 6 1 2 − 1 ) ( 1 6 1 − 1 6 1 2 − 1 ) = 1
( 1 6 1 + 1 6 1 2 − 1 ) − 1 = 1 6 1 − 1 6 1 2 − 1
Let a = 1 6 1 + 1 6 1 2 − 1 , b = 1 6 1 − 1 6 1 2 − 1 ⇒ a b = 1
N N 3 N 3 − 3 N N 3 N 3 N = = = = = = ≈ ≈ ≈ a 3 4 + b 3 4 a 4 + b 4 + 3 ( a b ) 3 4 ( a 3 4 + b 3 4 ) ( a 2 + b 2 ) 2 − 2 ( a b ) 2 + 3 N ( ( a + b ) 2 − 2 a b ) 2 − 2 + 3 N ( 3 2 2 2 − 2 ) 2 − 2 + 3 N ( 3 2 2 2 − 2 ) 2 − 2 ( 3 2 2 2 − 2 ) 2 ( 3 2 2 2 ) 2 3 2 2 3 4 ≈ 2 2 0 7 . 0 2 8
Trial and error shows that N = 2 2 0 7 satisfy the equation. The answer is 2 0 7