# Appending hexadecimal string for MD5 algorithm

I followed the algorithm step by step in MD5 Algorithm but in the first loop of the 64 loops, I got my A as 10325476, B as 35E4F8C4D, C as EFCDAB89 and D as 98BADCFE in hexadecimal. A, b, C and D will be appended in the end to become the md5 checksum. This makes me confuse as in the first loop, the length of hexadecimal is already 9. I believe that by continuing the algorithm, B will become bigger and bigger. This means that I will not get exactly 32 hexadecimal number in the end as 8+9+8+8!=32 Am I on the correct path or my hexadecimal length of B is indicating that I am calculating wrongly?

In MD5, the addition performed on $B$ at each round is modulo $2^{32}$ (that is, only the 32 low-order bits of the result are kept); same for the additions involving $A$, $F$, $K$ and $M$ at each round, and the four additions on $A$, $B$, $C$ and $D$ at the end of the 64 rounds.
Therefore, $A$, $B$, $C$ and $D$ all remain within 32 bits. At the end of the hashing, the concatenation of their representation as 32 bits forms the final hash, which is thus exactly 128-bit.