Question

let x be the least number divisible by 13,such that when x is divided by 4,5,6,7,8and12 the remainder in each case is 2 .find sum digits of x​

Answer 1

Answer:

8=2×2×2 , 12=2×2×3 , 16=2×2×2×2 , 20=2×2×5

LCM =2×2×2×2×3×5=240

We are to find out a lowest number which is divisible by 13 , and when 1 subtracted from it divisible by 240.

If we multiply 13 by those numbers whose units digit is 7 , then only we get 1 in the unit place of the product , and then only subtracting 1 we will get a number whose unit place digit is 0 , which creats posibility of divisibility by 240

13×7=91 , 13×17=221 , 13×27=351, 13×37=481

481–1=480 , 480÷240=2

So lowest number is 481 which is divisible by 13 , and if divided by 8,12,16,20 it will leave 1 as a remainder . ans