The problem states that when a number x is divided by 30, 45, or 54, the remainder is always 21. This means that x - 21 must be a multiple of 30, 45, and 54. To find the least value of x, we n
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Step 1: Understand the problem.
The problem states that when a number $x$ is divided by 30, 45, or 54, the remainder is always 21. This means that $x - 21$ must be a multiple of 30, 45, and 54. To find the least value of $x$, we need to find the least common multiple (LCM) of 30, 45, and 54.
Step 2: Find the prime factorization of each number.
• For 30:
$$30 = 2 \times 15 = 2 \times 3 \times 5$$
• For 45:
$$45 = 3 \times 15 = 3 \times 3 \times 5 = 3^2 \times 5$$
• For 54:
$$54 = 2 \times 27 = 2 \times 3 \times 9 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3$$
Step 3: Calculate the LCM of 30, 45, and 54.
To find the LCM, take the highest power of each prime factor present in any of the numbers:
• Highest power of 2: $2^1$
• Highest power of 3: $3^3$
• Highest power of 5: $5^1$
$$ \text{LCM}(30, 45, 54) = 2^1 \times 3^3 \times 5^1 = 2 \times 27 \times 5 = 270 $$
Step 4: Determine the value of $x$.
We know that $x - 21$ is the LCM.
$$x - 21 = 270$$
Add 21 to both sides to find $x$:
$$x = 270 + 21$$
$$x = 291$$
The least value of the number $x$ is $\boxed{\textbf{291}}$.
