To solve this problem, let's represent the number as \( N \) such that it satisfies the given conditions when divided by 5, 7, and 8. We are given:

• When \( N \) is divided by 5, the remainder is 2. Thus, \( N \equiv 2 \ (\text{mod} \ 5) \), which can be expressed as \( N = 5k + 2 \) for some integer \( k \).
• When \( N \) is divided by 7, the remainder is 3. Thus, \( N \equiv 3 \ (\text{mod} \ 7) \), or \( N = 7m + 3 \) for some integer \( m \).
• When \( N \) is divided by 8, the remainder is 4. Thus, \( N \equiv 4 \ (\text{mod} \ 8) \), or \( N = 8n + 4 \) for some integer \( n \).

To find such an \( N \), we'll apply the method of successive substitutions using the Chinese Remainder Theorem. Accordingly, we find a number \( N \) that satisfies all conditions:

Step 1: Start with the condition \( N \equiv 4 \ (\text{mod} \ 8) \). This implies that \( N = 8a + 4 \).
Step 2: Apply for \( 8a + 4 \equiv 3 \ (\text{mod} \ 7) \): 
\(8a + 4 \equiv 3 \ (\text{mod} \ 7)\)
Simplifying, we get \(8a \equiv -1 \equiv 6 \ (\text{mod} \ 7)\) because \(-1 \equiv 6 \) under modulo 7.
\( 8 \equiv 1 \ (\text{mod} \ 7) \), thus: 
\(a \equiv 6 \ (\text{mod} \ 7)\) 
Therefore, \( a = 7b + 6 \) for some integer \( b \).

Step 3: Substitute back \( a \) in \( N = 8a + 4 \): 
\(N = 8(7b + 6) + 4 = 56b + 48 + 4 = 56b + 52\)
Step 4: Apply \( 56b + 52 \equiv 2 \ (\text{mod} \ 5) \): 
Simplifying: \(56b + 52 \equiv 2 \ (\text{mod} \ 5) \Rightarrow 56 \equiv 1 \ (\text{mod} \ 5), 52 \equiv 2 \ (\text{mod} \ 5)\) 
\( 56b + 52 \equiv b + 2 \equiv 2 \ (\text{mod} \ 5) \Rightarrow b \equiv 0 \ (\text{mod} \ 5) \)

Therefore, \( b = 5c \) for some integer \( c \). Substitute back in the expression for \( N \): \(N = 56(5c) + 52 = 280c + 52.\) Verify with options provided:

Check for \( N = 437 \):

1. Dividing 437 by 8 gives a remainder of 4, satisfying \( N \equiv 4 (\text{mod} \ 8) \).
2. Dividing 437 by 7 gives a remainder of 3, satisfying \( N \equiv 3 (\text{mod} \ 7) \).
3. Dividing 437 by 5 gives a remainder of 2, satisfying \( N \equiv 2 (\text{mod} \ 5) \).

Thus, the correct number that matches all conditions is 437.