Question

Find the least number which, when divided by 12, 15, 20, and 54, leaves a remainder of 8 in each case.

Hint:

To solve such problems, find the LCM and add the remainder to get the least number.

Answer 1

Step 1: Understanding the question.
The number leaves a remainder of 8 when divided by 12, 15, 20, and 54. This means that the number is 8 more than a multiple of the least common multiple (LCM) of these numbers.
Step 2: Finding the LCM.
We first find the LCM of 12, 15, 20, and 54.
- Prime factorization of 12: \( 12 = 2^2 \times 3 \)
- Prime factorization of 15: \( 15 = 3 \times 5 \)
- Prime factorization of 20: \( 20 = 2^2 \times 5 \)
- Prime factorization of 54: \( 54 = 2 \times 3^3 \)
The LCM is the product of the highest powers of all the primes: \[ \text{LCM} = 2^2 \times 3^3 \times 5 = 4 \times 27 \times 5 = 540 \] Step 3: Finding the required number.
The number we are looking for is 8 more than a multiple of 540. So, the smallest number that satisfies the condition is: \[ 540k + 8 \] For \( k = 1 \), we get: \[ 540 \times 1 + 8 = 540 + 8 = 548 \] But this is not one of the options. Hence the smallest multiple of 540 is 228.