question_answer

Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.

Answer:

It is clear that the least number divisible by 6, 15 and 18 is their LCM. Therefore, the required number must by 5 more than their LCM.                                                                   Now the LCM of 6, 15, 18

2	6, 15, 18
3	3,  15, 9
3	1, 5, 3
5	1, 5, 1
1	1, 1, 1

                                                                                                                                                LCM \[=3\times 2\times 3\times 5=90\] Thus, the required number = 90 + 5 = 95