Let-f-be-a-one-to-one-function-from-the-set-of-natural-numbers-to-itself-such-that-f-mn-f-m-f-n-for-all-natural-numbers-m-and-n-What-is-the-least-possible-value-of-f-999-

Question Number 19978 by Tinkutara last updated on 19/Aug/17

Let f be a one - to - one function from

the set of natural numbers to itself

such that f (m n) = f (m) f (n) for all

natural numbers m and n . What is the

least possible value of f (999) ?

Answered by mrW1 last updated on 20/Aug/17

Every natural number N can be written

as the product of a serial of prime

numbers :

N = p_{1}^{q_{1}} \times p_{2}^{q_{2}} \times \dots \times p_{k}^{q_{k}}

p = prime numbers 1, 2, 3, 5, 7 \dots .

since f (mn) = f (m) f (n)

f (N) = {[f (p_{1})]}^{q_{1}} \times {[f (p_{2})]}^{q_{2}} \times \dots \times {[f (p_{k})]}^{q_{k}}

for f (N) to be a one to one funtion, it is

to say that f (p_{i}) with i = 1, 2, \dots \infty must

be different .

f (1) = f (1 \times 1) = f (1) \times f (1)

\Rightarrow f (1) = 1

i . e . only f (1) should be 1, f (p_{i}) with i > 2

can be freely choosen .

With N = 999 = 1 \times 3^{3} \times 37

\Rightarrow f (999) = f (1) \times {[f (3)]}^{3} \times f (37)

to keep f (999) as small as possible we

can choose

f (3) = 2 and f (37) = 3

\Rightarrow f (999) = 1 \times 2^{3} \times 3 = 24

Commented by Tinkutara last updated on 20/Aug/17

Thank you very much Sir!