Question

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

Answer 1

LCM of 8, 9, 10, 15, and 20 are given by

2	8, 9, 10, 15, 20
2	4, 9, 5, 15, 10
5	2, 9, 5, 15, 5
3	2, 9, 1, 3, 1
	2, 3, 1, 1, 1

LCM = 2 × 2 × 5 × 3 × 2 × 3

= 20 × 6 × 3

= 120 × 3

= 360

The smallest number = 360 + 5 = 365

Hence, 365 is the smallest number which when divided by 8, 9, 10, 15, and 20 gives a remainder of 5 every time.