Finding the Smallest Natural Number with a Common Remainder

The problem asks for the smallest natural number that, when divided by 3, 4, 6, and 7, always leaves a remainder of 2. Let's break down how to solve this type of problem.

When a number leaves the same remainder $R$ when divided by several different numbers (say, $D_1, D_2, D_3, \dots$), it means that if we subtract $R$ from the number, the result will be perfectly divisible by $D_1, D_2, D_3, \dots$. In other words, the number minus the remainder is a multiple of each of the divisors.

The smallest number that is a multiple of several numbers is their Least Common Multiple (LCM). So, the number we are looking for, minus the remainder, must be a multiple of the LCM of the divisors. The smallest such positive multiple is the LCM itself.

Therefore, the smallest number that satisfies the condition is the LCM of the divisors plus the common remainder.

Step 1: Finding the LCM of the Divisors

The divisors given are 3, 4, 6, and 7. We need to find the Least Common Multiple (LCM) of these numbers. We can find the LCM using prime factorization.

• Prime factorization of 3: 3
• Prime factorization of 4: $2 \times 2 = 2^2$
• Prime factorization of 6: $2 \times 3$
• Prime factorization of 7: 7

To find the LCM, we take the highest power of each prime factor that appears in any of the factorizations:

• The highest power of 2 is $2^2$.
• The highest power of 3 is $3^1$.
• The highest power of 7 is $7^1$.

Now, multiply these highest powers together:

$\text{LCM}(3, 4, 6, 7) = 2^2 \times 3 \times 7 = 4 \times 3 \times 7 = 12 \times 7 = 84$.

The LCM of 3, 4, 6, and 7 is 84. This is the smallest number that is divisible by 3, 4, 6, and 7 without any remainder.

Step 2: Calculating the Smallest Natural Number

The problem states that the natural number leaves a remainder of 2 in each case when divided by 3, 4, 6, and 7. Since the LCM, 84, is perfectly divisible by these numbers, adding the remainder 2 to the LCM will give us the smallest number that leaves this remainder.

Smallest number = LCM(3, 4, 6, 7) + Remainder

Smallest number = $84 + 2 = 86$.

So, the smallest natural number that satisfies the condition is 86.

Verifying the Result

Let's check if 86 leaves a remainder of 2 when divided by 3, 4, 6, and 7:

• $86 \div 3$: $86 = 3 \times 28 + 2$. Remainder is 2.
• $86 \div 4$: $86 = 4 \times 21 + 2$. Remainder is 2.
• $86 \div 6$: $86 = 6 \times 14 + 2$. Remainder is 2.
• $86 \div 7$: $86 = 7 \times 12 + 2$. Remainder is 2.

The calculations confirm that 86 leaves a remainder of 2 in each case. Since 86 is a natural number (a positive integer) and it's calculated using the smallest possible multiple (the LCM), it is the smallest such natural number.

Revision Table: Key Concepts

Additional Information: Numbers with Common Remainders

Problems involving finding a number that leaves a specific remainder when divided by multiple divisors are common in number theory. If a number $N$ leaves a remainder $R$ when divided by $D_1, D_2, \dots, D_k$, then $N - R$ is divisible by each of $D_1, D_2, \dots, D_k$. This means $N - R$ is a common multiple of $D_1, D_2, \dots, D_k$.

The common multiples of a set of numbers are the multiples of their LCM. So, $N - R$ must be a multiple of $\text{LCM}(D_1, D_2, \dots, D_k)$.

The possible values for $N - R$ are $k \times \text{LCM}(D_1, D_2, \dots, D_k)$, where $k$ is any integer.

So, possible values for $N$ are $k \times \text{LCM}(D_1, D_2, \dots, D_k) + R$.

If we are looking for the smallest positive integer $N$, we usually take the smallest positive multiple of the LCM, which is the LCM itself (when $k=1$). If $R \ge 0$ and $R$ is less than all divisors, then $k=1$ gives the smallest positive number, $N = \text{LCM}(D_1, D_2, \dots, D_k) + R$. However, if $R$ is such that $\text{LCM}(D_1, \dots) + R$ is not positive (which is not the case here as LCM is positive and R=2), or if $R$ was negative, we might need to consider $k=0, 1, 2, \dots$ to find the smallest positive $N$. In this problem, the remainder is 2 and the divisors are positive, so $84+2=86$ is positive and is indeed the smallest natural number.