Manual States:
In this case, the use of the routine 'random_message3' requires variable 1 to contain the number of the middle message of a group of three.
9880 NAME random_message3:RANDOM 0:DIVIDE 0,16385:VAR_ADD 1,0:PRINT_GEN 0:STOP
Line 74: It's already open.
Line 75: It is open already.
Line 76: No, it's already open.
RANDOM Vn : A random number (range -32768 to +32767) is loaded into Vn.
DIVIDE Vn,#m : The value in Vn is divided by #m.
VAR_ADD Va,Vb : The contents of Va are added to those of Vb.
So 75 is already known to us, and passed in via variable 1 on var_add
Another example chooses from five messages (middle message is number 15)
2350 RANDOM 0:DIVIDE 0,10923:ADD 0,15:PRINT_GEN 0:RETurn
using 10932 returns a choice of five messages 13, 14, 15, 16, 17
So if I want say, a range of ten, what would my 16385 or 10932 value be? I don't understand how this figure is worked out? I know what my message numbers are, say 10 in a row starting at 45 to 55, I'd pass 50 either directly or with a var, but it's the division figure I'm stuck on.

I should have paid more attention to Johnny Ball
