2018 AMC 12B Problem #18

Algebra Level 3

A function f f is defined recursively by f ( 1 ) = f ( 2 ) = 1 f(1) = f(2 )= 1 and

f ( n ) = f ( n − 1 ) − f ( n − 2 ) + n f(n) = f(n - 1) - f(n - 2) + n

for all integers n ≥ 3. n \geq 3. What is f ( 2018 ) f(2018) ?

This problem is part of the 2018 AMC 12B .
2019 2018 2016 2017 2020

Steven Yuan by Steven Yuan , 20, USA

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1 solution

Vedant Saini
Dec 10, 2018

f ( 1 ) = 1 \ f ( 2 ) = 1 f ( 3 ) = 3 f ( 4 ) = 6 f ( 5 ) = 8 f ( 6 ) = 8 f ( 7 ) = 7 \ f ( 8 ) = 7 f ( 9 ) = 9 f ( 10 ) = 12 f ( 11 ) = 14 f ( 12 ) = 14 f ( 13 ) = 13 \ f ( 14 ) = 13 f(1) = 1\ \\ \color{#20A900} f(2) = 1 \\ f(3) = 3 \\ f(4) = 6 \\ f(5) = 8 \\ f(6) = 8 \\ f(7) = 7\ \\ \color{#20A900}f(8) = 7 \\ f(9) = 9 \\ f(10) =12 \\ f(11) = 14 \\ f(12) = 14 \\ f(13) = 13\ \\ \color{#20A900}f(14) = 13

Note that f ( 6 k + 2 ) = ( 6 k + 2 ) − 1 f(6k + 2) = (6k + 2) - 1 (as in g r e e n \color{#20A900}green ) and

2018 = 6 ⋅ k + 2 2018 = 6 \cdot\ k + 2 (where k k is equal to 336 336 )

So, f ( 2018 ) = 2018 − 1 = 2017 f(2018) = 2018 - 1 = 2017

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