Find the smallest number such that
(i) it leaves a remainder 2 when divided by 3 ;
(ii) it leaves a remainder 3 when divided by 5 ;
(iii) it leaves a remainder 5 when divided by 7 .
+85 8 is the smallest number that satisfies (i) and (ii).
next, 8+(3*5)x = 5 (mod7), 8+15x = 5 (mod 7), and 8+x = 5 (mod 7), it's clear than 8+4 = 12 so x = 4
Thus the smallest x can be is x = 4 (mod 7) and x = 4.
The answer is 8 + 15(4) = 68, the smallest number that satisfies the three conditions.
