Question

Let $d$ be a function taking the positive integers to the nonnegative integers such that $d(p) = 1$ for any prime $p,$ and
d(ab) = b \cdot d(a) + a \cdot d(b)
for all positive integers $a$ and $b.$ Find the number of positive integers $n \le 10$ such that $d(n) = 1.$

Answer 1

To solve the problem, we need to analyze the function \( d(n) \) where \( d(p) = 1 \) for any prime \( p \) and \( d(ab) = b \cdot d(a) + a \cdot d(b) \) for all positive integers \( a \) and \( b \).

First, let's evaluate \( d(n) \) for different positive integers \( n \):

1. For \( n = 1 \):
- We know that \( 1 \) has no prime factors, so we need to deduce \( d(1) \). Using \( d(ab) \) where \( a = b = 1 \), we have:
\[
d(1 \cdot 1) = 1 \cdot d(1) + 1 \cdot d(1) \implies d(1) = 2d(1) \implies d(1) = 0.
\]

2. For \( n = 2 \):
- \( d(2) \) is directly given by \( d \) for primes, so \( d(2) = 1 \).

3. For \( n = 3 \):
- Similarly, since \( 3 \) is prime, \( d(3) = 1 \).

4. For \( n = 4 \):
- We can express \( 4 \) as \( 2 \cdot 2 \). So we use the property:
\[
d(4) = d(2 \cdot 2) = 2 \cdot d(2) + 2 \cdot d(2) = 2 \cdot 1 + 2 \cdot 1 = 4.
\]

5. For \( n = 5 \):
- Since \( 5 \) is prime, \( d(5) = 1 \).

6. For \( n = 6 \):
- We can express \( 6 \) as \( 2 \cdot 3 \):
\[
d(6) = d(2 \cdot 3) = 3 \cdot d(2) + 2 \cdot d(3) = 3 \cdot 1 + 2 \cdot 1 = 5.
\]

7. For \( n = 7 \):
- Since \( 7 \) is prime, \( d(7) = 1 \).

8. For \( n = 8 \):
- We can express \( 8 = 2 \cdot 4 \):
\[
d(8) = d(2 \cdot 4) = 4 \cdot d(2) + 2 \cdot d(4) = 4 \cdot 1 + 2 \cdot 4 = 4 + 8 = 12.
\]

9. For \( n = 9 \):
- We can express \( 9 = 3 \cdot 3 \):
\[
d(9) = d(3 \cdot 3) = 3 \cdot d(3) + 3 \cdot d(3) = 3 \cdot 1 + 3 \cdot 1 = 6.
\]

10. For \( n = 10 \):
- We can express \( 10 = 2 \cdot 5 \):
\[
d(10) = d(2 \cdot 5) = 5 \cdot d(2) + 2 \cdot d(5) = 5 \cdot 1 + 2 \cdot 1 = 5 + 2 = 7.
\]

Now we summarize the results:
- \( d(1) = 0 \)
- \( d(2) = 1 \)
- \( d(3) = 1 \)
- \( d(4) = 4 \)
- \( d(5) = 1 \)
- \( d(6) = 5 \)
- \( d(7) = 1 \)
- \( d(8) = 12 \)
- \( d(9) = 6 \)
- \( d(10) = 7 \)

Finally, we see that \( d(n) = 1 \) for \( n = 2, 3, 5, 7 \). Thus, the number of positive integers \( n \le 10 \) such that \( d(n) = 1 \) is:

\[
\boxed{4}.
\]