Question

What is the least number which when divided by 21,28,36 and 45 leaves remainder 5 in each case?

Answer 1

Answer:

Step-by-step explanation:

Let’s see straight through it…. All three numbers will only divide a number when all of them are factors of it. So it’s obvious we need to find the LCM.

16 = 2^4

28 = 2^2 X 7

40 = 2^3 X 5

LCM = 2^4 X 7 X 5 (Taking the highest power of the nos.)

LCM = 560

So 560 Is the smallest number which the above three nos divide . But what is requested is that the smallest number which leaves the remainder 5.

So if we increase 5 from 560 , every time these three nos will leave a remainder as 5. (Isn’t that obvious) :)

Hence the no. will be 560+ 5 = 565

Answer 2

LCM of 21,28,36, 45= 560
If we devide 560 by 21,28,36, 45we get 0 in remiainder
So by adding 5 were reminder 5
So required no. = 565