What is the least number which when divided by 8, 12 and 16 leaves 3 as the remainder in each case, but when divided by 7 leaves no any remainder?

Answer

To solve the problem, we need to find the least number that satisfies two conditions:

1. When divided by 8, 12, and 16, it leaves a remainder of 3.
2. When divided by 7, it leaves no remainder.

Let's break down the solution step by step:

### Step 1: Find the LCM of 8, 12, and 16
To find the least number that leaves a remainder of 3 when divided by 8, 12, and 16, we first need to calculate the least common multiple (LCM) of these three numbers.

- The prime factorization of the numbers:
- 8 = 2^3
- 12 = 2^2 × 3
- 16 = 2^4

- The LCM is found by taking the highest power of each prime factor:
- For 2, the highest power is 2^4 (from 16).
- For 3, the highest power is 3^1 (from 12).

Thus, the LCM is:
\[ \text{LCM} = 2^4 \times 3^1 = 16 \times 3 = 48 \]

### Step 2: Adjust for the remainder
Since we want a number that, when divided by 8, 12, and 16, leaves a remainder of 3, we can express this as:
\[ N = 48k + 3 \]
where \( k \) is a non-negative integer.

### Step 3: Find \( N \) such that \( N \) is divisible by 7
Now, we need to find the smallest \( N \) that is also divisible by 7:
\[ N = 48k + 3 \]
We want:
\[ 48k + 3 \equiv 0 \mod 7 \]

Calculating \( 48 \mod 7 \):
\[ 48 \div 7 = 6 \quad \text{(remainder 6)} \]
So, \( 48 \equiv 6 \mod 7 \).

Now substituting this into our equation:
\[ 6k + 3 \equiv 0 \mod 7 \]
Subtracting 3 from both sides gives:
\[ 6k \equiv -3 \mod 7 \]
Since \(-3\) is equivalent to \(4\) modulo \(7\):
\[ 6k \equiv 4 \mod 7 \]

### Step 4: Solve for \( k \)
To solve for \( k \), we need to find the multiplicative inverse of \( 6 \mod 7\). The inverse of \( 6 \) modulo \( 7 \) is \( 6 \) itself, since:
\[ 6 \times 6 = 36 \equiv 1 \mod 7 \]

Now multiplying both sides of \( 6k \equiv 4 \) by \( 6 \):
\[ k \equiv 6 \times 4 \mod 7 \]
\[ k \equiv 24 \mod 7 \]
Calculating \( 24 \mod 7 \):
\[ 24 \div 7 = 3 \quad \text{(remainder 3)} \]
So, \( k \equiv 3 \mod 7 \).

### Step 5: Calculate \( N \)
The smallest non-negative \( k \) that satisfies this is \( k = 3 \). Now substituting back to find \( N \):
\[ N = 48 \times 3 + 3 = 144 + 3 = 147 \]

### Final Answer
The least number which when divided by 8, 12, and 16 leaves a remainder of 3, and when divided by 7 leaves no remainder is:
\[ \boxed{147} \]