Why does Java's hashCode() in String use 31 as a multiplier?

Question

Per the Java documentation, the hash code for a String object is computed as:

s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1] 

using int arithmetic, where s[i] is the ith character of the string, n is the length of the string, and ^ indicates exponentiation.

Why is 31 used as a multiplier?

I understand that the multiplier should be a relatively large prime number. So why not 29, or 37, or even 97?

Answer 1

According to Joshua Bloch's Effective Java (a book that can't be recommended enough, and which I bought thanks to continual mentions on stackoverflow):

The value 31 was chosen because it is an odd prime. If it were even and the multiplication overflowed, information would be lost, as multiplication by 2 is equivalent to shifting. The advantage of using a prime is less clear, but it is traditional. A nice property of 31 is that the multiplication can be replaced by a shift and a subtraction for better performance: 31 * i == (i << 5) - i. Modern VMs do this sort of optimization automatically.

(from Chapter 3, Item 9: Always override hashcode when you override equals, page 48)