Final answer:
(b) n = 3, which states that when 7 consecutive integers are each raised to the third power and summed, the total will be divisible by 7.
Step-by-step explanation:
The correct answer is option (b) n = 3. When dealing with the properties of integers and divisibility rules, raising integers to the third power (or any odd power) and summing them keeps the symmetry that is necessary for the sum to be divisible by 7. Let's consider 7 consecutive integers where x is the smallest.
These integers can be represented as x, x+1, x+2, x+3, x+4, x+5, and x+6. When these are raised to the third power and added together, the cubed terms with the same absolute distance from x+3 (the middle term) will pair up and their contributions will be symmetrical around x+3.
For example, (x+3)3 will be the central term in our expression, and the pairs (x)3+(x+6)3, (x+1)3+(x+5)3, and (x+2)3+(x+4)3 will all sum to multiples of 7. Hence, when you add all these terms together, the result is divisible by 7, fulfilling the stated condition.
In order for the total of the set of 7 consecutive integers, each raised to the power of n, to be divisible by 7, n must be a multiple of 3. This is because when we raise each integer to the power of n, the remainder when dividing by 7 will repeat every 3rd power.
For example, if we take the set of consecutive integers {1, 2, 3, 4, 5, 6, 7} and raise each number to the power of 2, we get {1, 4, 9, 16, 25, 36, 49}. The sum of these numbers is 140, which is divisible by 7.