Question

find the largest and smallest number of 5 digit which is divisible by 24,32 and 40
please solve it ASAP​

Answer 1

Step-by-step explanation:

First find the L.C.M. of 24,32 and 40.

2 |24,32,40

2 |12,16,20

2 |6,8,10

|3,4,5 =>2×2×2×3×4×5 =>480.

Largest five digit number is 99,999. Divide 99,999 by 480 we get remainder as 159 so the largest five digit number which is divisible by 24,32 and 40 is 99,840. Smallest five digit number is 10,000. When we divide 10,000 by 480 we get 400 as remainder. Hence the smallest five digit number can be obtained by adding 80 to 10,000 so 10,080 is the smallest five digit number which is divisible by 24,32 and40.Hope it helps you.