The least number that should be added to 2,397 so that the sum is exactly divisible by 3, 4, 5 and 6 ?

Options

A. a) 6
B. b) 4
C. c) 5
D. d) 3

Answer

To solve the problem of finding the least number that should be added to 2,397 so that the sum is exactly divisible by 3, 4, 5, and 6, we can follow these steps:

### Step 1: Find the Least Common Multiple (LCM)
First, we need to find the LCM of the numbers 3, 4, 5, and 6.

- The prime factorization of each number is:
- 3 = 3^1
- 4 = 2^2
- 5 = 5^1
- 6 = 2^1 * 3^1

To find the LCM, we take the highest power of each prime number:
- For 2: the highest power is 2^2 (from 4)
- For 3: the highest power is 3^1 (from 3 or 6)
- For 5: the highest power is 5^1 (from 5)

Thus, the LCM is:
\[ \text{LCM} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \]

**Hint:** To find the LCM, always consider the highest power of each prime factor from the numbers involved.

### Step 2: Check the Current Number
Next, we check the number 2,397 to see how much we need to add to make it divisible by 60.

### Step 3: Find the Remainder
We calculate the remainder when 2,397 is divided by 60:
\[ 2397 \div 60 = 39 \quad \text{(whole number)} \]
Calculating \( 60 \times 39 = 2340 \).

Now, we find the remainder:
\[ 2397 - 2340 = 57 \]

**Hint:** To find the remainder, subtract the largest multiple of the divisor (60) that is less than the number (2397).

### Step 4: Calculate the Required Addition
To make 2,397 divisible by 60, we need to add enough to reach the next multiple of 60. Since the remainder is 57, we need to add:
\[ 60 - 57 = 3 \]

### Conclusion
Thus, the least number that should be added to 2,397 to make it exactly divisible by 3, 4, 5, and 6 is **3**.

**Final Answer:** 3