Final answer:
For the expression (2x^2 + 1)/3, the three possible remainders when divided by 3 are 0, 1, and 2 since the remainder must be less than the divisor according to the division algorithm.
Step-by-step explanation:
To determine how many remainders the expression (2x^2 + 1)/3 can have for all natural numbers x, we can apply the division algorithm. The division algorithm states that given any integer a, and any positive integer b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.
In the context of this problem, 'a' corresponds to the numerator 2x^2 + 1, and 'b' is 3. Thus, we have:
2x^2 + 1 = 3q + r
Now, we need to find how many different values 'r' (the remainder) can take. Since 'b' is 3, the remainder 'r' can only be 0, 1, or 2, according to the division algorithm. We cannot have a remainder equal to or larger than the divisor.
Let's consider all possibilities:
- If x is a multiple of 3 (say 3s), then 2x^2 + 1 is also divisible by 3, thus r = 0.
- If x is one more than a multiple of 3 (3p + 1), then (2x^2 + 1) mod 3 is equal to 2.
- If x is two more than a multiple of 3 (3d + 2), then (2x^2 + 1) mod 3 is equal to 1.
Therefore, for the natural number x, the expression (2x^2 + 1)/3 can have three different remainders: 0, 1, or 2.