Find the sum of digit of a smallest number which when divided by 16, 24, 30 and 36 leaves remainder 8, 16, 22 and 28 respectively?

Options

A. 9
B. 10
C. 13
D. 11

Answer

To solve the problem, we need to find the smallest number that leaves specific remainders when divided by 16, 24, 30, and 36. The remainders are 8, 16, 22, and 28 respectively.

### Step-by-step Solution:

1. **Understanding the Problem**:
We need to find a number \( x \) such that:
- \( x \mod 16 = 8 \)
- \( x \mod 24 = 16 \)
- \( x \mod 30 = 22 \)
- \( x \mod 36 = 28 \)

2. **Rewriting the Conditions**:
We can rewrite these conditions in terms of \( x \):
- \( x = 16k + 8 \) for some integer \( k \)
- \( x = 24m + 16 \) for some integer \( m \)
- \( x = 30n + 22 \) for some integer \( n \)
- \( x = 36p + 28 \) for some integer \( p \)

3. **Finding a Common Form**:
To find a common number that satisfies all these conditions, we can express \( x \) in a more manageable form. We can adjust the conditions to find a common multiple:
- \( x - 8 \equiv 0 \mod 16 \)
- \( x - 16 \equiv 0 \mod 24 \)
- \( x - 22 \equiv 0 \mod 30 \)
- \( x - 28 \equiv 0 \mod 36 \)

This means \( x - 8 \) must be a common multiple of the divisors (16, 24, 30, 36).

4. **Finding the LCM**:
We need to calculate the least common multiple (LCM) of the numbers 16, 24, 30, and 36.
- Prime factorization:
- \( 16 = 2^4 \)
- \( 24 = 2^3 \times 3^1 \)
- \( 30 = 2^1 \times 3^1 \times 5^1 \)
- \( 36 = 2^2 \times 3^2 \)

The LCM takes the highest power of each prime:
- \( LCM = 2^4 \times 3^2 \times 5^1 = 16 \times 9 \times 5 = 720 \)

5. **Finding the Smallest Number**:
Now, since \( x - 8 \) is a multiple of 720, we can express \( x \) as:
\[ x = 720k + 8 \]
To find the smallest positive \( x \), we set \( k = 1 \):
\[ x = 720 \times 1 + 8 = 728 \]

6. **Calculating the Sum of Digits**:
Now, we find the sum of the digits of 728:
\[ 7 + 2 + 8 = 17 \]

### Final Answer:
The sum of the digits of the smallest number is **17**.