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

1. 

Let N be the number of coins of Alex

2a+1=N
3b+1=N
4c+1=N
5d+1=N
6e+1=N
7f+1=N

so N-1 is a multiple of 2, of 3, of 4, of 5, of 6 and of 7

so N-1 has factors 7*6*5*2*X, where X is another factor. 

2.
There is no need to write N-1 as 7*6*5*3*2*4, since being a multiple of 6, means being a multiple of 2, and of 3 as well. Also, no need to write *4, since 6 has already *2, but we need one more 2, that why we write 7*6*5*2*X

(7*6*5*2=420 is the LCM, least common multiple of 2,3,4,5,6,7)

3.

N-1=420*X. 

X cannot be 2 or more since N would become larger that 800.

X must be 1, so N-1=420, which means N=421 is the number of coins

He is right i did this thing before