Which is the smallest number, which on dividing by 18, 24, 30 and 42 leaves remainder as 1?

Options

A. 2510
B. 2512
C. 2521
D. 2601

Answer

To find the smallest number that leaves a remainder of 1 when divided by 18, 24, 30, and 42, we can follow these steps:

### Step 1: Find the Least Common Multiple (LCM)
We need to find the LCM of the numbers 18, 24, 30, and 42.

- **Factorization of each number:**
- 18 = 2 × 3²
- 24 = 2³ × 3
- 30 = 2 × 3 × 5
- 42 = 2 × 3 × 7

### Step 2: Determine the highest power of each prime factor
Now, we take the highest power of each prime factor present in the factorizations:

- For 2: The highest power is 2³ (from 24)
- For 3: The highest power is 3² (from 18)
- For 5: The highest power is 5¹ (from 30)
- For 7: The highest power is 7¹ (from 42)

### Step 3: Calculate the LCM
Now we can calculate the LCM by multiplying these highest powers together:

\[ \text{LCM} = 2³ × 3² × 5¹ × 7¹ \]

Calculating this step-by-step:

- \(2³ = 8\)
- \(3² = 9\)
- \(5¹ = 5\)
- \(7¹ = 7\)

Now, multiply these together:

\[ 8 × 9 = 72 \]
\[ 72 × 5 = 360 \]
\[ 360 × 7 = 2520 \]

So, the LCM of 18, 24, 30, and 42 is 2520.

### Step 4: Adjust for the remainder
Since we need a number that leaves a remainder of 1 when divided by these numbers, we add 1 to the LCM:

\[ 2520 + 1 = 2521 \]

### Final Answer
Thus, the smallest number that leaves a remainder of 1 when divided by 18, 24, 30, and 42 is **2521**.

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