Find the least number which when divided by 16, 18 and 20 leaves a remainder 4 in each case, but is completely divisible by 7.

Options

A. 2884
B. 2256
C. 865
D. 3332

Answer

To find the least number which when divided by 16, 18, and 20 leaves a remainder of 4 in each case, and is completely divisible by 7, we can follow these steps:

### Step 1: Find the Least Common Multiple (LCM) of 16, 18, and 20
To find the LCM, we need to determine the prime factorization of each number:
- 16 = 2^4
- 18 = 2^1 × 3^2
- 20 = 2^2 × 5^1

Now, we take the highest power of each prime factor:
- For 2: max(4, 1, 2) = 4 → 2^4
- For 3: max(0, 2, 0) = 2 → 3^2
- For 5: max(0, 0, 1) = 1 → 5^1

Thus, the LCM is:
\[ \text{LCM} = 2^4 × 3^2 × 5^1 = 16 × 9 × 5 = 720 \]

### Step 2: Adjust for the Remainder
Since we want a number that leaves a remainder of 4 when divided by 16, 18, and 20, we can express our desired number as:
\[ N = \text{LCM} + 4 = 720 + 4 = 724 \]

### Step 3: Ensure Divisibility by 7
Now we need to find the smallest multiple of the LCM (720) that, when added to 4, is divisible by 7. We can express this as:
\[ N = 720k + 4 \]
where \( k \) is a positive integer.

We need to find the smallest \( k \) such that:
\[ (720k + 4) \mod 7 = 0 \]

### Step 4: Calculate \( 720 \mod 7 \)
First, we find \( 720 \mod 7 \):
\[ 720 \div 7 = 102 \text{ remainder } 6 \]
So, \( 720 \equiv 6 \mod 7 \).

### Step 5: Set Up the Equation
Now we substitute this into our equation:
\[ (6k + 4) \mod 7 = 0 \]
This simplifies to:
\[ 6k + 4 \equiv 0 \mod 7 \]
\[ 6k \equiv -4 \mod 7 \]
Since \(-4 \equiv 3 \mod 7\), we have:
\[ 6k \equiv 3 \mod 7 \]

### Step 6: Solve for \( k \)
To solve for \( k \), we can multiply both sides by the modular inverse of 6 modulo 7. The inverse of 6 modulo 7 is 6 (since \( 6 \times 6 \equiv 1 \mod 7 \)):
\[ k \equiv 6 \times 3 \mod 7 \]
\[ k \equiv 18 \mod 7 \]
\[ k \equiv 4 \mod 7 \]

Thus, the smallest positive integer \( k \) is 4.

### Step 7: Calculate the Final Number
Now, substituting \( k = 4 \) back into our equation:
\[ N = 720 \times 4 + 4 = 2880 + 4 = 2884 \]

### Conclusion
The least number which when divided by 16, 18, and 20 leaves a remainder of 4, and is completely divisible by 7 is:
\[ \boxed{2884} \]