• # question_answer What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder of 2 but when divided by 13 leaves no remainder? A)  962                            B)  692          C)  269                D)  629

LCM of 3, 5, 6, 8, 10 and 12 =120. So, the required number is of the form $120\,\,k+2.$ Least value of $k$for which $(120k+2)$ is divisible by 13 is$k=8$. $\therefore$Required number$=(120\times 8+2)=962.$