Let N is the least number of 6 digits, which when divide by 4, 6, 10 and 15 leaves the remainder 2 in each case. Find the sum of digits in N.

Options

A. 3
B. 5
C. 4
D. 6

Answer

To solve the problem step by step, we will find the least 6-digit number \( N \) that leaves a remainder of 2 when divided by 4, 6, 10, and 15. Then, we will calculate the sum of the digits of \( N \).

### Step 1: Find the LCM of the divisors
First, we need to find the least common multiple (LCM) of the numbers 4, 6, 10, and 15.

- The prime factorization of each number is:
- \( 4 = 2^2 \)
- \( 6 = 2^1 \times 3^1 \)
- \( 10 = 2^1 \times 5^1 \)
- \( 15 = 3^1 \times 5^1 \)

- To find the LCM, we take the highest power of each prime that appears in these factorizations:
- For \( 2 \): the highest power is \( 2^2 \) (from 4)
- For \( 3 \): the highest power is \( 3^1 \) (from 6 and 15)
- For \( 5 \): the highest power is \( 5^1 \) (from 10 and 15)

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

### Step 2: Find the smallest 6-digit number
The smallest 6-digit number is \( 100000 \).

### Step 3: Adjust for the remainder
Since we need \( N \) to leave a remainder of 2 when divided by 4, 6, 10, and 15, we can express \( N \) as:
\[ N = k \times 60 + 2 \]
where \( k \) is an integer.

To find the smallest \( k \) such that \( N \) is a 6-digit number, we set up the inequality:
\[ k \times 60 + 2 \geq 100000 \]
Subtracting 2 from both sides gives:
\[ k \times 60 \geq 99998 \]
Dividing both sides by 60:
\[ k \geq \frac{99998}{60} \approx 1666.6333 \]
Since \( k \) must be an integer, we take \( k = 1667 \).

### Step 4: Calculate \( N \)
Now we can calculate \( N \):
\[ N = 1667 \times 60 + 2 = 100020 + 2 = 100022 \]

### Step 5: Find the sum of the digits of \( N \)
Now, we will find the sum of the digits of \( 100022 \):
\[ 1 + 0 + 0 + 0 + 2 + 2 = 5 \]

### Final Answer
The sum of the digits in \( N \) is \( 5 \).

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