Question

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

Hint:

Build the LCM two numbers at a time instead of factorising all four together: find the LCM of the first pair, then combine that result with the next number, and so on. Once you have the final LCM, remember that adding the given remainder to it gives the smallest number satisfying the condition.

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: \[ 540 \times 1 + 8 = 540 + 8 = 548 \] This is the smallest such number. Step 4: Conclusion.
The required number is 548.

Answer 2

Concept:

• The LCM of several numbers can be built up two at a time instead of factorising all of them together at once, which keeps each individual step simple.
• Once the LCM is known, a fixed remainder condition is handled by adding the remainder directly to the LCM, since the LCM itself is the smallest number divisible by every one of them with zero remainder.

Step 1: Find the LCM of the first two numbers.
LCM of 12 and 15: $12 = 2^2 \times 3$, $15 = 3 \times 5$, so LCM $= 2^2 \times 3 \times 5 = 60$.

Step 2: Bring in the third number using the LCM found so far.
LCM of 60 and 20: $20 = 2^2 \times 5$ already divides 60 exactly, so the LCM stays $60$.

Step 3: Bring in the fourth number the same way.
LCM of 60 and 54: $54 = 2 \times 3^3$. Combining the highest powers with 60 $= 2^2 \times 3 \times 5$ gives $2^2 \times 3^3 \times 5 = 540$.

Step 4: Add the fixed remainder to the LCM.
A number leaving remainder 8 with every one of 12, 15, 20, 54 is 8 more than a common multiple of all four, and the smallest such common multiple is the LCM itself, 540.
$540 + 8 = 548$

Final Answer: The required number is 548.