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 .

Guest Jul 27, 2022

#1**+8 **

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.

Doremy Jul 27, 2022