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- Feb 14, 2012

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(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 $f(67500)$.

- Thread starter anemone
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- #1

- Feb 14, 2012

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(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 $f(67500)$.

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- Mar 5, 2012

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Let me give it a try.

(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 $f(67500)$.

With (c) this yields:

$$f(2) + 2f(3) + f(5) - 3 = 5$$

$$f(2) + 2f(3) + f(5) = 8\qquad [1]$$

Neither f(2), nor f(3), nor f(5) can be 1.

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

It follows from (a) that:

$f(67500)=f(2^2\cdot 3^3\cdot 5^4)=2f(2)+3f(3)+4f(5) - 8=2\cdot 2 + 3\cdot 2 + 4\cdot 2 - 8 = 10. \qquad \blacksquare$

- Aug 30, 2012

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Clever!Let me give it a try.

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).

Combiningmy lemmawith [1] tells us that $f(2)=f(3)=f(5)=2$.

It follows from (a) that:

$f(67500)=f(2^2\cdot 3^3\cdot 5^4)=2f(2)+3f(3)+4f(5) - 8=2\cdot 2 + 3\cdot 2 + 4\cdot 2 - 8 = 10. \qquad \blacksquare$

-Dan

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- Feb 14, 2012

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Thank you for participating,Let me give it a try.

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).

Combiningmy lemmawith [1] tells us that $f(2)=f(3)=f(5)=2$.

It follows from (a) that:

$f(67500)=f(2^2\cdot 3^3\cdot 5^4)=2f(2)+3f(3)+4f(5) - 8=2\cdot 2 + 3\cdot 2 + 4\cdot 2 - 8 = 10. \qquad \blacksquare$

Indeed!Clever!

-Dan