A number x is divisible by 7 . When this number is divided by 8, 12 and 16 . It leaves as remainder 3 in each case. The least value of x is

Options

A. 148
B. 149
C. 150
D. 147

Answer

To solve the problem step by step, we need to find the least value of \( x \) that meets the given conditions.

### Step 1: Understand the conditions
The number \( x \) is:
- Divisible by 7
- Leaves a remainder of 3 when divided by 8, 12, and 16

### Step 2: Set up the equations
Since \( x \) leaves a remainder of 3 when divided by 8, 12, and 16, we can express \( x \) in the following form:
\[ x = LCM(8, 12, 16) \cdot k + 3 \]
where \( k \) is a positive integer.

### Step 3: Calculate the LCM
To find the least common multiple (LCM) of 8, 12, and 16:
- The prime factorization of 8 is \( 2^3 \)
- The prime factorization of 12 is \( 2^2 \cdot 3^1 \)
- The prime factorization of 16 is \( 2^4 \)

The LCM takes the highest power of each prime:
- For \( 2 \), the highest power is \( 2^4 \)
- For \( 3 \), the highest power is \( 3^1 \)

Thus, the LCM is:
\[ LCM(8, 12, 16) = 2^4 \cdot 3^1 = 16 \cdot 3 = 48 \]

### Step 4: Express \( x \)
Now we can express \( x \) as:
\[ x = 48k + 3 \]

### Step 5: Check divisibility by 7
We need \( x \) to be divisible by 7:
\[ 48k + 3 \equiv 0 \mod 7 \]
Calculating \( 48 \mod 7 \):
\[ 48 \div 7 = 6 \quad \text{(remainder 6)} \]
So, \( 48 \equiv 6 \mod 7 \). Therefore, we rewrite the equation:
\[ 6k + 3 \equiv 0 \mod 7 \]
Subtracting 3 from both sides:
\[ 6k \equiv -3 \mod 7 \]
This is equivalent to:
\[ 6k \equiv 4 \mod 7 \quad \text{(since -3 is equivalent to 4 in mod 7)} \]

### Step 6: Solve for \( k \)
To solve \( 6k \equiv 4 \mod 7 \), we can multiply both sides by the modular inverse of 6 modulo 7. The inverse of 6 modulo 7 is 6 (since \( 6 \cdot 6 \equiv 1 \mod 7 \)):
\[ k \equiv 6 \cdot 4 \mod 7 \]
Calculating:
\[ k \equiv 24 \mod 7 \quad \Rightarrow \quad k \equiv 3 \mod 7 \]
Thus, the smallest positive integer \( k \) is 3.

### Step 7: Calculate \( x \)
Substituting \( k = 3 \) back into the equation for \( x \):
\[ x = 48 \cdot 3 + 3 = 144 + 3 = 147 \]

### Final Answer
The least value of \( x \) is:
\[ \boxed{147} \]