To solve the problem, we need to find a positive integer \( x \) that satisfies the following congruences:
- \( x \equiv 1 \mod 5 \)
- \( x \equiv 4 \mod 8 \)
We'll solve for \( x \) that meets these criteria and then find \( y \) such that \( x + y \) is divisible by 40.
Step 1: Find \( x \)
From the first congruence, we have:
\[ x = 5k + 1 \quad \text{for some integer } k \]
Substituting this expression for \( x \) into the second congruence:
\[ 5k + 1 \equiv 4 \mod 8 \]
Subtracting 1 from both sides gives:
\[ 5k \equiv 3 \mod 8 \]
Now, we can find the multiplicative inverse of 5 modulo 8. The inverse \( m \) satisfies:
\[ 5m \equiv 1 \mod 8 \]
Testing values, we find:
- \( m = 1: 5 \times 1 = 5 \not\equiv 1 \mod 8 \)
- \( m = 2: 5 \times 2 = 10 \equiv 2 \mod 8 \)
- \( m = 3: 5 \times 3 = 15 \equiv 7 \mod 8 \)
- \( m = 4: 5 \times 4 = 20 \equiv 4 \mod 8 \)
- \( m = 5: 5 \times 5 = 25 \equiv 1 \mod 8 \) (found the inverse!)
Multiplying both sides of \( 5k \equiv 3 \mod 8 \) by 5 gives:
\[ 25k \equiv 15 \mod 8 \implies k \equiv 7 \mod 8 \]
Thus, \( k \) can be expressed as:
\[ k = 8m + 7 \quad \text{for some integer } m \]
Substituting back for \( x \):
\[ x = 5(8m + 7) + 1 = 40m + 35 + 1 = 40m + 36 \]
So we have:
\[ x \equiv 36 \mod 40 \]
The smallest positive integer \( x \) satisfying both congruences is \( x = 36 \).
Step 2: Find \( y \)
Now we need to find the smallest positive integer \( y \) such that \( x + y \) is divisible by 40:
\[ x + y \equiv 0 \mod 40 \]
Since \( x \equiv 36 \mod 40 \):
\[ 36 + y \equiv 0 \mod 40 \implies y \equiv -36 \equiv 4 \mod 40 \]
The smallest positive integer \( y \) satisfying this is:
\[ y = 4 \]
Conclusion
The smallest positive integer \( y \) such that \( (x + y) \) is divisible by 40 is:
\[ \boxed{4} \]