ElectricPavlov

Rate of SLOWER crane =  1 ship /   (x -10 )

Rate of FASTER crane =   1 ship / x 

Now for one ship with both working the time = 12 hrs:

    1 ship /  ( 1/(x-10)  +   1/x)  =  12  hr       <====     solve for 'x' 

1 = 12/(x-10)  + 12 / x

x^2 - 10x = 12x + 12x-120

x^2 -34x +120 = 0      Use Quadrtaic Formula  ( or factoring) to find x =   30     or 4 (extraneous)

                                         so one crane takes  30 hours      the other  takes  20  hours alone     

a)  each of the other 6 rows has 8 less 

          73 - 6 *8 = 25 in the top row

b)       \(\sum_{0}^{6}\)  73 - 8n  = 343 boxes    

The lowest prime in the list would be 11     400/11 = 36.4    so the biggest prime int the list will be 31    11*31 = 341

   so from 11 to 31   the list of primes is  11  13  17  19   23  29  31    

           11 x the other 6   is 6 possibilities 

               13 x  17, 19,23,29    is 4 more possibles

                      17 x 19,  23   is two more 

                          and that is all          12 possible products 

B = benedict's original # of cars

S = sam's cars =  B-130 

Now give away some cars 

B will now have     B -  3/4 ( B-130 )        and S will have   (B-130) - .4 ( B)        These two equations are now equal 

   B - 3/4 (B-130 )    =   ( B-130) - .4B    solve for B

   ,25B +  97.5  =  .6B - 130 

    227.5  =  .35 B 

B=650 cars at first    

Hmmmm....   I dunno....    to ME   digits   is plural...as in more than one digit....

   others may read this as you did which would make your answer correct....

So....this question is subject to interprotation.... making it a poorly worded problem !   

194  fits the bill