fix(hashes): resolve clippy manual_is_multiple_of warnings

Replace manual modulo checks with the cleaner is_multiple_of() method:
- hashes/src/hex.rs: Check for odd-length hex strings
- hashes/src/sha256.rs: Check block size alignment in from_midstate

🤖 Generated with [Claude Code](https://claude.ai/code)

Co-Authored-By: Claude <noreply@anthropic.com>

Author: PastaPastaPasta

hashes/src/hex.rs

@@ -86,7 +86,7 @@ impl<'a> HexIterator<'a> {
     ///
     /// If the input string is of odd length.
     pub fn new(s: &'a str) -> Result<HexIterator<'a>, Error> {
-        if s.len() % 2 != 0 {
+        if !s.len().is_multiple_of(2) {
             Err(Error::OddLengthString(s.len()))
         } else {
             Ok(HexIterator {

hashes/src/sha256.rs

@@ -404,7 +404,7 @@ impl HashEngine {
     ///
     /// If `length` is not a multiple of the block size.
     pub fn from_midstate(midstate: Midstate, length: usize) -> HashEngine {
-        assert!(length % BLOCK_SIZE == 0, "length is no multiple of the block size");
+        assert!(length.is_multiple_of(BLOCK_SIZE), "length is no multiple of the block size");
 
         let mut ret = [0; 8];
         for (ret_val, midstate_bytes) in ret.iter_mut().zip(midstate[..].chunks_exact(4)) {