 Hello and welcome to the session. In this session we discussed the following question which says, find the smallest number which when decreased by 5 is divisible by 1215, 20 and 27. Let's see the solution now. So we have to find the smallest number which when decreased by 5 would also be divisible by 1215, 20 and 27. We know that the smallest number divisible by 1215, 20 and 27 is their LCM. So let's first find out the LPM of 1215, 20 and 27. Now 2, 6 times is 12, 15 is not the divisible by 2 so we write 15 as it is. 2, 10 times is 20, 27 is not the divisible by 2 so we write 27 as it is. Then again consider 2, 2 3 times is 6, 15 is not divisible by 2 so we write 15 as it is. Now 5, 2 times is 10 so we write 5 here, 27 is not divisible by 2 so we write 27 here. Next consider 3, 3 1 times is 3, 3 5 times is 15 then 5 is not divisible by 3 so we write 5 as it is. Now 9, 3 times is 27 so we write 9 here. Next consider 3 again, 5 is not divisible by 3 so we write 5 as it is. Then 3, 3 times is 9. Now again 3, 3, 1 times is 3 since 5 is not divisible by 3 so we write 5 as it is. Now 5, 1 times is 5 so now LPM of 1215, 20 and 27 is the product of these numbers which is equal to 2 into 2 into 3 into 3 into 3 into 5 so we get LPM of 1215, 20 and 27 as 540 so we say that the smallest number divisible by 12, 15, 20 and 27 is 540 but we have to find the smallest number which when decreased by 5 is divisible by 12, 15, 20 and 27 so the required number would be equal to 540 plus 5 equal to 545. Hence we say the smallest number which when decreased by 5 is divisible by 12, 15, 20 and 27 is 545. So this number so 545 is our final answer. This completes the session. Hope you have understood the solution of this question.