Understanding the Problem: Number Divisible by 35 and 7 Leaving a Remainder of 2

In this article, we explore the process of finding a number that, when divided by 35 and 7, leaves a remainder of 2 in each case. We'll delve into the mathematical foundation, solution methods, and practical applications of this concept.

Mathematical Foundation

The problem at hand can be expressed using modular arithmetic. If a number x leaves a remainder of 2 when divided by 35 and 7, we can write:

x ≡ 2 (mod 3)

x ≡ 2 (mod 5)

x ≡ 2 (mod 7)

This can be alternatively expressed as:

x - 2 ≡ 0 (mod 3)

x - 2 ≡ 0 (mod 5)

x - 2 ≡ 0 (mod 7)

From here, we can deduce that x - 2 must be divisible by the least common multiple (LCM) of 3, 5, and 7.

Calculating the LCM

The least common multiple of 3, 5, and 7 is calculated as:

LCM(3, 5, 7) 3 × 5 × 7 105

Thus, we can rewrite the equation as:

x - 2 105k

Where k is any integer. Adding 2 to both sides gives us:

x 105k 2

Determining Suitable Values

To find suitable values for x, we can substitute different integer values for k:

If k 0, then x 105 × 0 2 2 If k 1, then x 105 × 1 2 107 If k 2, then x 105 × 2 2 212

The general solution for x is:

x 105k 2

For the smallest positive solution, we use k 1, giving us:

x 105 × 1 2 107

The boxed solution is:

boxed{107}

Additional Numbers

By increasing k, we can find other numbers that satisfy the criteria:

For k 2, x 105 × 2 2 212 For k 3, x 105 × 3 2 317 For k 4, x 105 × 4 2 422

A convenient way to find such numbers is to note that any multiple of 105 plus 2 will satisfy the condition:

x 105k 2

Further Exploration

The LCM of 35 and 7 is also 105. Adding 2 to this, we get:

357 105, 105 2 107

Therefore, whether we take the LCM of 3 and 5, 3 and 7, 5 and 7, or all three together, the number that satisfies the condition is:

357 105, 105 2 107

Some other numbers between 1 and 1000 that satisfy the condition are:

2, 107, 212, 317, 422, 527, 632, 737, 842, 947

This demonstrates the rich network of numbers that satisfy the given conditions.

Conclusion

In summary, we have found that the smallest number that is divisible by 35 and 7 and leaves a remainder of 2 in each case is 107. This number can be generated by the general formula x 105k 2. Understanding and applying these concepts can aid in solving similar problems in number theory and modular arithmetic.