johng said:Fun little problem.
First p=2017 is a prime congruent to 1 mod 4; i.e (p-1)/2 is even. Let $x=2\times4\times\cdots \times p-1$ and $y=1\times3\times5\cdots\times p-2$. Now there are (p-1)/2 factors in x. Write x in reverse order: $x=p-1\times p-3\times p-5\cdots p-(p-2)$. So mod p we see that $x=(-1)^{(p-1)/2}1\times3\times5\cdots\times p-2=y$. So x - y is 0 mod p. Notice the result is true for any prime that is 1 mod 4.
The mathematical expression is (2*4*6*8*...*2016) - (1*3*5*7*...*2015).
To find the remainder, we can use the modulus operator (%) which gives the remainder when one number is divided by another. In this case, we would divide the result of the expression by 2017 and use the remainder as the answer.
Yes, it can be solved using a shortcut method called the Chinese Remainder Theorem. This method involves breaking down the problem into smaller parts and finding the remainder for each part, then combining them to find the final remainder.
The number 2017 is a prime number, which means it is only divisible by 1 and itself. This makes it a useful number in mathematics, especially in problems involving remainders.
Yes, this problem can be generalized for any series of even and odd numbers. The formula would be (a1*a2*a3*...*an)-(b1*b2*b3*...*bm), where a1 to an are even numbers and b1 to bm are odd numbers.