Be 3 fast

Note $B$ has more than one trailing zero (from the factors of 10,20) but $A$ has exactly one trailing zero (from the factors of $2,5$ ; since the other factors are prime they cannot be divisible by $2$ or $5$ ). Therefore, the sum $A+B$ has exactly $\boxed{1}$ trailing zero.

$A$ has only one trailing zero, while $B$ has a lot of. The second step I took is realizing that $B>A$ , so $A+B$ has only one trailing zero.

The product of the first 100 primes is... 11 356 769 871 639 834 366 628 078 343 660 801 213 021 002 603 808 017 113 484 083 456 481 393 624 183 355 161 423 988 601 739 578 097 269 175 036 999 016 380 044 894 643 879 365 114 584 536 377 648 421 251 919 478 851 653 433 974 523 196 310 641 702 368 516 916 764 541 545 146 381 989 396 087 448 510

Take the last 3 digits (510) + 100 = 610 <- 1 Trailing Zero