Finding the Smallest 4-Digit Number

The problem asks for the smallest 4-digit number that leaves a remainder of 1 when divided by 2, 3, and 5.

Understanding the Conditions

• The number must be a 4-digit number (i.e., greater than or equal to 1000).
• When the number is divided by 2, the remainder is 1.
• When the number is divided by 3, the remainder is 1.
• When the number is divided by 5, the remainder is 1.

This means the number, let's call it N, can be expressed as:

N = (Multiple of 2) + 1

N = (Multiple of 3) + 1

N = (Multiple of 5) + 1

Therefore, N - 1 must be a multiple of 2, 3, and 5.

Calculating the Least Common Multiple (LCM)

To find a number that is a multiple of 2, 3, and 5, we need to find their Least Common Multiple (LCM).

The numbers 2, 3, and 5 are prime numbers. The LCM of prime numbers is their product.

$ \text{LCM}(2, 3, 5) = 2 \times 3 \times 5 = 30 $

So, N - 1 must be a multiple of 30. This can be written as:

N - 1 = 30k

Which means:

N = 30k + 1

where k is an integer.

Finding the Smallest 4-Digit Number

We are looking for the smallest 4-digit number, which means N must be greater than or equal to 1000.

$ 30k + 1 \ge 1000 $

Subtract 1 from both sides:

$ 30k \ge 999 $

Divide by 30:

$ k \ge \frac{999}{30} $

$ k \ge 33.3 $

Since k must be an integer, the smallest integer value for k that satisfies this condition is 34.

Determining the Number

Now, substitute k = 34 back into the equation for N:

$ N = (30 \times 34) + 1 $

$ N = 1020 + 1 $

$ N = 1021 $

The smallest 4-digit number that satisfies the given conditions is 1021.