Finding the Smallest Natural Number with a Specific Remainder

The question asks for the smallest natural number that, when divided by 4, 5, 6, or 7, consistently leaves a remainder of 3. Let the required number be \(N\).

Understanding the Remainder Condition

When a number \(N\) is divided by another number \(d\) and leaves a remainder \(r\), it means that \(N = q \times d + r\), where \(q\) is the quotient. Another way to think about this is that \(N - r\) is perfectly divisible by \(d\).

In this problem, the number \(N\) leaves a remainder of 3 when divided by 4, 5, 6, and 7. This implies the following conditions:

• \(N\) divided by 4 leaves a remainder of 3, so \(N - 3\) is divisible by 4.
• \(N\) divided by 5 leaves a remainder of 3, so \(N - 3\) is divisible by 5.
• \(N\) divided by 6 leaves a remainder of 3, so \(N - 3\) is divisible by 6.
• \(N\) divided by 7 leaves a remainder of 3, so \(N - 3\) is divisible by 7.

Therefore, the number \(N - 3\) is a common multiple of 4, 5, 6, and 7.

Using Least Common Multiple (LCM)

We are looking for the smallest such natural number \(N\). This means we need to find the smallest possible value for \(N - 3\). The smallest positive common multiple of a set of numbers is their Least Common Multiple (LCM).

So, the smallest possible value for \(N - 3\) is the LCM of 4, 5, 6, and 7.

Calculating the LCM of 4, 5, 6, and 7

To find the LCM, we first find the prime factorization of each number:

• Prime factors of 4: \(4 = 2 \times 2 = 2^2\)
• Prime factors of 5: \(5 = 5^1\) (5 is a prime number)
• Prime factors of 6: \(6 = 2 \times 3\)
• Prime factors of 7: \(7 = 7^1\) (7 is a prime number)

The LCM is found by taking the highest power of all prime factors that appear in any of the numbers:

The prime factors involved are 2, 3, 5, and 7.

• Highest power of 2: \(2^2\) (from 4)
• Highest power of 3: \(3^1\) (from 6)
• Highest power of 5: \(5^1\) (from 5)
• Highest power of 7: \(7^1\) (from 7)

LCM(4, 5, 6, 7) = \(2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7\)

Calculating the product:

\(4 \times 3 = 12\)

\(12 \times 5 = 60\)

\(60 \times 7 = 420\)

So, the LCM of 4, 5, 6, and 7 is 420.

This means the smallest value for \(N - 3\) is 420.

Finding the Smallest Number N

We established that \(N - 3 = \text{LCM}(4, 5, 6, 7)\). Since LCM(4, 5, 6, 7) = 420, we have:

\(N - 3 = 420\)

To find \(N\), we add 3 to 420:

\(N = 420 + 3\)

\(N = 423\)

Verification

Let's check if dividing 423 by 4, 5, 6, and 7 leaves a remainder of 3 in each case:

• 423 ÷ 4: \(423 = 105 \times 4 + 3\). Remainder is 3. Correct.
• 423 ÷ 5: \(423 = 84 \times 5 + 3\). Remainder is 3. Correct.
• 423 ÷ 6: \(423 = 70 \times 6 + 3\). Remainder is 3. Correct.
• 423 ÷ 7: \(423 = 60 \times 7 + 3\). Remainder is 3. Correct.

The number 423 satisfies all the conditions. Since we used the Least Common Multiple, 420 is the smallest number that is a common multiple of 4, 5, 6, and 7. Therefore, 420 + 3 = 423 is the smallest number that leaves a remainder of 3 when divided by these numbers.

Comparing with Options

Let's look at the given options:

Based on our calculation and verification, the smallest number is 423.

Revision Table: Key Concepts

Additional Information: Generalizing Remainder Problems

Problems involving finding a number that leaves the same remainder when divided by multiple divisors can be solved using the LCM concept.

If a number \(N\) leaves a remainder \(r\) when divided by numbers \(d_1, d_2, \dots, d_k\), then \(N - r\) is a common multiple of \(d_1, d_2, \dots, d_k\).

The smallest such positive number \(N\) is given by:

\[N = \text{LCM}(d_1, d_2, \dots, d_k) + r\]

In our problem, \(d_1=4, d_2=5, d_3=6, d_4=7\), and \(r=3\). We found LCM(4, 5, 6, 7) = 420. So the smallest number is \(420 + 3 = 423\).

Other numbers that satisfy the condition would be of the form \(k \times \text{LCM}(4, 5, 6, 7) + 3\), where \(k\) is a natural number (1, 2, 3, ...). For \(k=1\), we get 423. For \(k=2\), we get \(2 \times 420 + 3 = 840 + 3 = 843\), which was one of the options.

This shows why 423 is the smallest such natural number.