The four digit smallest positive number which when divided by 4, 5, 6 or 7, it always leaves the remainder as 3 :

Options

A. 1000
B. 1257
C. 1263
D. 1683

Answer

To find the smallest four-digit positive number that leaves a remainder of 3 when divided by 4, 5, 6, or 7, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 4 = 3 \)
- \( N \mod 5 = 3 \)
- \( N \mod 6 = 3 \)
- \( N \mod 7 = 3 \)

This means that \( N - 3 \) must be divisible by 4, 5, 6, and 7.

### Step 2: Find the Least Common Multiple (LCM)
To solve for \( N - 3 \), we need to find the least common multiple (LCM) of the divisors 4, 5, 6, and 7.

- The prime factorization of each number is:
- \( 4 = 2^2 \)
- \( 5 = 5^1 \)
- \( 6 = 2^1 \times 3^1 \)
- \( 7 = 7^1 \)

The LCM is found by taking the highest power of each prime:
- \( 2^2 \) from 4
- \( 3^1 \) from 6
- \( 5^1 \) from 5
- \( 7^1 \) from 7

Thus, the LCM is:
\[ \text{LCM} = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 \]

Calculating this step-by-step:
- \( 4 \times 3 = 12 \)
- \( 12 \times 5 = 60 \)
- \( 60 \times 7 = 420 \)

So, \( \text{LCM}(4, 5, 6, 7) = 420 \).

### Step 3: Set Up the Equation
Now, we can express \( N \) as:
\[ N = 420k + 3 \]
where \( k \) is a non-negative integer.

### Step 4: Find the Smallest Four-Digit Number
We need \( N \) to be a four-digit number:
\[ 1000 \leq 420k + 3 < 10000 \]
Subtracting 3 from all parts:
\[ 997 \leq 420k < 9997 \]
Dividing by 420:
\[ \frac{997}{420} \leq k < \frac{9997}{420} \]
Calculating the bounds:
- \( \frac{997}{420} \approx 2.37 \) (so \( k \geq 3 \))
- \( \frac{9997}{420} \approx 23.80 \) (so \( k \leq 23 \))

Thus, \( k \) can take integer values from 3 to 23.

### Step 5: Calculate the Smallest \( N \)
Using the smallest integer \( k = 3 \):
\[ N = 420 \times 3 + 3 = 1260 + 3 = 1263 \]

### Conclusion
The smallest four-digit positive number which when divided by 4, 5, 6, or 7 always leaves a remainder of 3 is **1263**.

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