Question

let us find the least number divisible by 13 such that when that number is divided by 8,12,16 and 20 it leaves 1 as remainder all the cases

Answer 1

first of all we have to take the LCM of 8,12,16,20
LCM of 8,12,16,20=240
then we have to find term lower than 240 which is divisible by 13 which is 234

now we can write it like this
234x+6x+1 we write 1 here because we want remainder 1
so 234 is divisible by 13 so now we have to find the value for 6x+1 which is divisible by 13
6×2+1=13 is the value which is divisible by 13
now this value is written as follows to find our number
240×13+1=1921
1921 is the no. which is divisible by 13 and have remainder 1 when divide by 8,12,16 and 20
I hope I help you