Question

Find the smallest number which when divided by 8,9,10,15,20 gives a remainder of 5 every time.

Answer 1

Here, the smallest number for division is LCM of 8, 9, 10,15 and 20. 

8 = 2 x 2 x 2 

9 = 3 x 3 

10 = 2 x 5 

15 = 3 x 5 

20 = 2 x 2 x 5 

LCM of given numbers = 2 x 2 x 2 x 3 x 3 x 5 

= 360 

∴ Required, smallest number = LCM + Remainder 

= 360 + 5

= 365 

∴ The required smallest number is 365.