Find the least number which when divided by 5,7 and 8 leaves 3 as the remainder in each case.

Options

A. 283
B. 78
C. 578
D. 57

Answer

To find the least number which, when divided by 5, 7, and 8, leaves a remainder of 3 in each case, we can follow these steps:

### Step 1: Understand the problem
We need to find a number \( N \) such that:
- \( N \mod 5 = 3 \)
- \( N \mod 7 = 3 \)
- \( N \mod 8 = 3 \)

This means that \( N \) can be expressed in the form:
\[ N = 5k + 3 \]
\[ N = 7m + 3 \]
\[ N = 8n + 3 \]
for some integers \( k, m, n \).

### Step 2: Rewrite the equations
We can rewrite the equations without the remainder:
- \( N - 3 \) must be divisible by 5, 7, and 8.
Let \( M = N - 3 \). Then:
\[ M \mod 5 = 0 \]
\[ M \mod 7 = 0 \]
\[ M \mod 8 = 0 \]

### Step 3: Find the least common multiple (LCM)
We need to find the least common multiple of 5, 7, and 8 to determine the smallest \( M \):
- The prime factorization of the numbers:
- \( 5 = 5^1 \)
- \( 7 = 7^1 \)
- \( 8 = 2^3 \)

The LCM is found by taking the highest power of each prime:
\[ \text{LCM}(5, 7, 8) = 5^1 \times 7^1 \times 2^3 = 5 \times 7 \times 8 = 280 \]

### Step 4: Calculate \( M \)
The smallest \( M \) that satisfies the conditions is 280. Therefore:
\[ M = 280 \]

### Step 5: Find \( N \)
Now, we can find \( N \):
\[ N = M + 3 = 280 + 3 = 283 \]

### Conclusion
The least number which when divided by 5, 7, and 8 leaves a remainder of 3 in each case is:
\[ \boxed{283} \]