# Find f(67500)

#### anemone

##### MHB POTW Director
Staff member
Given that the function $f(x)$ is defined on the set of natural numbers, taking values from the natural numbers, and that it satisfies the following conditions:

(a) $f(xy)=f(x)+f(y)-1$ for any $x,y \in N$.

(b) the equality $f(x)=1$ is true for finitely many numbers.

(c) $f(90) = 5$

Find $f(67500)$.

#### MarkFL

Staff member
I have edited anemone's post to write $f(90)=5$ instead of $f(90)=6$. I helped her modify this problem from a past Olympiad problem, and I inadvertently gave her the wrong value. My apologies to everyone.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Given that the function $f(x)$ is defined on the set of natural numbers, taking values from the natural numbers, and that it satisfies the following conditions:

(a) $f(xy)=f(x)+f(y)-1$ for any $x,y \in N$.

(b) the equality $f(x)=1$ is true for finitely many numbers.

(c) $f(90) = 5$

Find $f(67500)$.
Let me give it a try.

From (a) we find that $f(90) = f(2\cdot 3^2\cdot 5) = f(2) + 2f(3) + f(5) - 3$.
With (c) this yields:
$$f(2) + 2f(3) + f(5) - 3 = 5$$
$$f(2) + 2f(3) + f(5) = 8\qquad [1]$$

My lemma
Neither f(2), nor f(3), nor f(5) can be 1.
Proof
Suppose one of them is 1, say f(2), then $f(2^k) = kf(2) - k + 1=1$.
That means that infinitely many numbers x have f(x)=1.
This is a contradiction with (b).

Combining my lemma with [1] tells us that $f(2)=f(3)=f(5)=2$.

It follows from (a) that:
$f(67500)=f(2^2\cdot 3^3\cdot 5^4)=2f(2)+3f(3)+4f(5) - 8=2\cdot 2 + 3\cdot 2 + 4\cdot 2 - 8 = 10. \qquad \blacksquare$

#### topsquark

##### Well-known member
MHB Math Helper
Let me give it a try.

From (a) we find that $f(90) = f(2\cdot 3^2\cdot 5) = f(2) + 2f(3) + f(5) - 3$.
With (c) this yields:
$$f(2) + 2f(3) + f(5) - 3 = 5$$
$$f(2) + 2f(3) + f(5) = 8\qquad [1]$$

My lemma
Neither f(2), nor f(3), nor f(5) can be 1.
Proof
Suppose one of them is 1, say f(2), then $f(2^k) = kf(2) - k + 1=1$.
That means that infinitely many numbers x have f(x)=1.
This is a contradiction with (b).

Combining my lemma with [1] tells us that $f(2)=f(3)=f(5)=2$.

It follows from (a) that:
$f(67500)=f(2^2\cdot 3^3\cdot 5^4)=2f(2)+3f(3)+4f(5) - 8=2\cdot 2 + 3\cdot 2 + 4\cdot 2 - 8 = 10. \qquad \blacksquare$
Clever!

-Dan

#### anemone

##### MHB POTW Director
Staff member
Let me give it a try.

From (a) we find that $f(90) = f(2\cdot 3^2\cdot 5) = f(2) + 2f(3) + f(5) - 3$.
With (c) this yields:
$$f(2) + 2f(3) + f(5) - 3 = 5$$
$$f(2) + 2f(3) + f(5) = 8\qquad [1]$$

My lemma
Neither f(2), nor f(3), nor f(5) can be 1.
Proof
Suppose one of them is 1, say f(2), then $f(2^k) = kf(2) - k + 1=1$.
That means that infinitely many numbers x have f(x)=1.
This is a contradiction with (b).

Combining my lemma with [1] tells us that $f(2)=f(3)=f(5)=2$.

It follows from (a) that:
$f(67500)=f(2^2\cdot 3^3\cdot 5^4)=2f(2)+3f(3)+4f(5) - 8=2\cdot 2 + 3\cdot 2 + 4\cdot 2 - 8 = 10. \qquad \blacksquare$
Thank you for participating, I like Serena! And your answer is correct!

Clever!

-Dan
Indeed!