Let f defined as f (3n) = n + f (3n – 3), where n is a positive integer greater than 1. If f (3n) = 1 when n=1, then find last digit of f (312)
Let f defined as f (3n) = n + f (3n – 3), where n is a positive integer greater than 1. If f (3n) = 1 when n=1, then find last digit of f (312)
Option 1 -
8
Option 2 -
6
Option 3 -
4
Option 4 -
0
Asked by Shiksha User
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1 Answer
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Correct Option - 1
Detailed Solution:f (3n) – f (3n– 3) = n
n = 1
f (3) – f (0) = 1
f (0) = 0
n = 2
f (3n) – f (3) = 2 + 3 + 4+…………….+ n
f (3n) = 1 + 2 + 3 +…………….+ n
=
f (1312) – f (3 × 311)
Hence, remainder = 8
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