Alex was counting his coins by 2s. because one coin was left over he counted them by 3s. again there was one coin left over so he counted by 4s the 5s the 6s and finally 7s. each time there was one left over. knowing that alex did not have more than 800 coins exactly how many coins did he have

(My process of thinking)

Well, we know that however many coins he does have, x, is not more than 800 AND cannot be divided by 2, 3, 4, 5, 6, or 7. So, I need to find a number that can divide into 800 but the rest cannot.

x mod 3=1, x mod 4 = 1, x mod 5 = 1 and x mod 7 = 1.

Note that if x mod 3 = 1 and x mod 4 = 1 then x must be of the form 12a+1. So (12a+1) mod 2 = 1

and (12a+1) mod 6 = 1. That is you need not have the equations for 2 and 6.

To satisfy the 4 equations above x must be of the form (3*4*5*7) b+1 = 420b+1. Knowing that x <800 we get that b=1 and so x=421. Alex had 421 coins!