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TMUA 2021 · Paper 1 · Question 5 of 20

TMUA 2021 Paper 1 Question 5

Algebra and functions — Functional equations · a piecewise multiplicative rule. Try it first; the answer and a full worked solution are below.

TMUA 2021 · Paper 1Algebra and functionsFunctional equations · a piecewise multiplicative rule6 options
The function f is such that, for all positive integers m and n:
  • f(mn) = f(m)f(n) if mn is a multiple of 3
  • f(mn) = mn if mn is not a multiple of 3

Given that f(9) + f(16) − f(24) = 0, what is the value of f(3)?

  1. A83
  2. B2√2
  3. C3
  4. D165
  5. E3√2
  6. F4
Show the answer and worked solution
answer · F
  1. A83
  2. B2√2
  3. C3
  4. D165
  5. E3√2
  6. F4
Split each argument into a product and apply the right branch. 9 = 3 × 3 is a multiple of 3, so f(9) = f(3)2. 16 = 2 × 8 is not a multiple of 3, so f(16) = 16; the same reasoning gives f(8) = 8. For 24 = 3 × 8, which is a multiple of 3, f(24) = f(3)f(8) = 8f(3). The condition becomes f(3)2 − 8f(3) + 16 = 0, that is (f(3) − 4)2 = 0, so f(3) = 4.