Question

find the least number which when divided by 24, 36, 45and 54 leaves remainder 5 in each case​

Answer 1

Answer:

Therefore, the smallest number which when divided by 24, 26 and 54 gives a remainder of 5 each time is 221

Answer 2

Answer:

221

Step-by-step explanation:

Prime factorization of

24 = 2 × 2 × 2 × 3

36 = 2 × 2 × 3 × 3

54 = 2 × 3 × 3 × 3

So the required LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216

The smallest number which is exactly divisible by 24, 36 and 54 is 216

In order to get remainder as 5

Required smallest number = 216 + 5 = 221

Therefore, the smallest number which when divided by 24, 26 and 54 gives a remainder of 5 each time is 221.