Yes. this is doing to infamous division by 10 using bit shifts as you will find in hacker's delight: http://www.hackersdelight.org/divcMore.pdf.

You may notice that when you divide num by 10 you get a quotient q and a remainder r:

    num = q*10 + r

Once you get q, you can solve for r as

    r = num - q*10

So this is how you get r and q:

    q = (num >> 1) + (num >> 2);
    q = q + (q >> 4);
    q = q + (q >> 8);
    q = q + (q >> 16);
    q = q >> 3;
    r = num - q*10;
    q = q + ((r + 6) >> 4)    

voila!