Plus/minus everything

In the set of $2018$ integers, $1009$ are even, and $1009$ are odd.

$\pm2\pm4\pm6\pm\ldots\pm2016\pm2018$ results in an even number.

$\pm1\pm3\pm5\pm\ldots\pm2015\pm2017$ results in an odd number, as there are an odd number of these exclusively odd numbers.

Adding the even number to the odd number results in an odd number therefore the answer must be odd, and the only odd option is $\boxed{1001}$ .

Starting with $-1$ and alternating signs gives $-1 + 2 - 3 + 4 - \dots - 2017 + 2018 = (-1 + 2) + (-3 + 4) + \dots + (-2017 + 2018) = \frac{2018}{2} = 1009$ . Changing the $+4$ to a $-4$ makes $1009 - 4 - 4 = \boxed{1001}$ .

Since there are an odd number of odd numbers, the sum must be odd, so the even choices cannot be obtained.

Edit: 2037173 has been removed from the choices.

First of all, $1+2+3+ ... + 2018=2037171$ , so clearly, 2037173 is impossible.

Now, note that changing any of the + signs to - signs will not affect the parity of the answer. Therefore, since 2037171 is odd, only odd numbers can be obtained. The only odd number left is 1001.

Notice that there are $1009$ odd numbers in the set $\{ 1,2,\ldots,2018 \}$ . This is because every odd number is of the form $2k - 1$ , so there is bijection between sets $\{ 1,2\ldots, 1009 \}$ and $\{ 1,3,\ldots, 2017 \}$ given by $k\mapsto 2k-1$ . Thus,

$\pm 1 \pm 2 \pm 3\pm \ldots \pm 2017 \pm 2018 \equiv \pm 1 \pm 0 \pm 1 \pm \ldots \pm 1 \pm 0 \equiv \underbrace{1+1+\ldots +1}_{1009} \equiv 1009 \equiv 1 \pmod 2$

which proves that it is impossible to obtain even numbers, leaving $1001$ as the only option.

To prove that we can obtain $1001$ , notice that the same reasoning as above gives us that there are $1009$ pairs $\{ 2k-1, 2k\}$ in the set $\{ 1,2,\ldots,2018 \}$ , so

$-1+2-3+4-\ldots -2017+2018 = (-1+2) + (-3+4) + \ldots +(-2017+2018) = \underbrace{1+1+\ldots +1}_{1009} = 1009.$

If we switch signs of any pair $\{2k-1,2k\}$ in the above sum, the sum will decrease by $2$ , since $(-(2k-1) + 2k) - ((2k-1) - 2k) = 2$ . Thus, pick $4$ distinct pairs $\{ 2k-1,2k\}$ for sign switch in the above sum to obtain $1001$ .

Only odd numbers can be obtained because the sum of all the numbers is odd, and changing the $+$ and $-$ signs doesn't change the parity of the answer. 1001 is the only odd answer choice, so it has to be the right one.

If you're not satisfied with that answer, and you want proof 1001 is obtainable, then

$-1+2-3+4-5...-2017+2018=(-1+2)+(-3+4)+(-5+6)...+(-2017+2018)=2018/2=1009$ .

Changing the $+4$ into $-4$ gives $1009-4-4=1001$

The sum or substraction of an even and an odd number is odd. We have 1009 even and 1009 odd numbers, which means and addition or substraction of and odd number of numbers which is necessarily odd. If you put -1 and +1002, you can group the rest of the numbers by four numbers with a zero sum : -2+3+4-5 for example. You do it from 2 to 1001 (1000 numbers, multiple of 4) and from 1003 to 2018 (1016 numbers, multiple of 4)

This problem reminds me of an anecdote about Carl Friedrich Gauss. When he went to basic school, one day his math teacher wanted a break and asked his pupils to add up the integers from 1 to 100. He thought that it would keep them busy for a while. But only a few moments later, young Gauss shouted: "5050". He had found that the formula $(n+1)\cdot \frac{n} {2}$ did the job easily. Applied to our problem, in case of the max sum (all signs +) we see the two factors n+1=2019 and n/2 = 1009 are both odd, so the max sum is odd. Changing the sign of any of the integers i will reduce the sum by the even number 2i to a sum that is an odd number again. Thus, only 1001 is the correct result of the problem.

We group the numbers into pairs. {(1,2) ,(3,4) ,(4,5) ... (1017,1018)}. This list has size 1009. For each pair we subtract the first from the second. This gives a new set of numbers, {1, 1, 1... 1}, with same size 1009. The sum of these are 1009.

Now we see that the closest numbers are 1001 and 1000. So we go back to the pairs and see if we can alter a single value to subtract either 8 or 9 from the initial sum. In the pair (3,4) we can see having minus in front of both 3 and 4 will give -7. This is 8 less than 1 and will alter the result to be 1001.

I think that it would be a good idea to call the sum +1+2+....+2017+2018 = S is an odd number (1/2 X 2018X2019 = 2,037,171). And note that subtracting the first number 1 and adding the rest gives (-1) + 2 + ... + 2,017 + 2,018. And this can be written [2 x (-1) + (1 + 2 + ..... + 2,017 + 2,018)] or [2 x (-1) + S] Which is odd since the sum of all the numbers has been shown to be odd, and you are just subtracting and even number from this. So the answer has to be odd. As 3 of the answers are even, they must be wrong. BUT SURELY WE HAVE STILL NOT SHOWN THAT THE ODD ANSWER THAT IS GIVEN (1001) CAN BE CORRECT! We need the numbers that are added to sum to 1,019,086 and the numbers that are subtracted to add to 1,018,085. And note that the sum of all the numbers up to and including 1,426 equals .5 x 1426 x 1427 = 1,017,451. Then adding in addition the number 1,635 gets us to 1,019,086. Which is one of the ways (though by no means the unique way) of getting the answer of 1001 by adding all the numbers as I have said and then subtracting all the other numbers. (Note I do not think that all odd numbers between 1 and 2.037,171 will be able to be obtained, though I'm not certain about this! Perhaps someone else will investigate this, I think it will be interesting, I will investigate myself!!). Regards, David

Arranging the numbers like this, (-1+2)+(-3+4)+(-5+6)+...+(2017-2018)= ? There are 2018/2=1009 pairs of numbers that result 1. So 1009x1 is equal 1009.

Look: (+/- doesn't matter, the number will keep odd/even)

》 Even+even+even=even> even-even-even=even 》 Odd+odd+odd= odd> odd-odd-odd=odd 》Odd+even= odd > odd-even=odd 》Even+odd= odd > even-odd=odd

We have 1009 numbers even and so odd. Then, if we separe odd numbers and even ones: 》Odd sum = odd (1009 is odd and odd plus odd, odd times= odd number) 》Even sum= even (even numbers always sum even) 》Even sum + odd sum =odd number (a great number) 》Even sum - odd sum= odd number. If we do this step the result will be exactly 1009.

As you can see, there is no way to rearrange the numbers to get an even number. To get to the answer, you separe the number +- 1 and change the signal of the 2 first pairs (1,2) and (3,4)

(-1+2)+(-3+4)+(-5+6)+...+(2017-2018)

(-1+2)+(-3+4)... -> 1 (-2-3-4) -> 1 + (-9) -> (-8)+(-5+6)+...+(2017-2018) = 1001

Assume that we put negative sign for all odd numbers and taking the sum gives $1009$ (i.e., $Even-Odd$ = $\frac {2018}{2}$ = $1009$ )which is closer to the options 1000 and 1001. Now putting a negative on the number eight gives you the sum $1001$ .

Only odd answer 1001.

I first put the numbers into pairs.

So you have:

1 + 2018 = 2019

2 + 2017 = 2019

... and so on.

2018/2 = 1009.

Therefore there are 1009 of these pairs. These pairs can cancel each other out.

+1 + 2018 - 2 - 2017 = 0

But there is an odd number of pairs. If everything else equals to 0 then the surplus pair must be the one that gives us the actual value.

What do we know about this pair? Well, firstly it must give

a + b = 2019

Now it is only a matter of finding which numbers that add up to 2019 can also minus to give 4, 1000, 1001 or 2018. They have to have that precise difference.

(a + 4) + a = 2019 gives us a non-integer value for a, meaning it cannot be part of the pair, since the values are all integers.

So does

(a + 1000) + a = 2019

and

(a + 2018) + a = 2019

1001 has to be the answer.

(a + 1001) + a = 2019

a = 509

Here, a is an integer. Therefore 1001 is the only possible answer.

Any four consecutive integers can be added/subtracted to get zero. We have 2018 numbers in sequence, which mod 4 gives us two. Two consecutive integers give us one as a difference, so our result will always be odd (swapping adding a number for subtracting, changes the result by twice the number, thus keeping parity).

The only odd number in the choice of answers is 1001, so that must be our only choice.

Let’s argue by using the parity of the sum/difference on the left hand side.
We have 1009 even numbers, and 1009 odd numbers. Sum and difference of two odd and/or even numbers results in an even number. Since we have an odd number of odd numbers, it follows that the result from finite addition/subtraction will be odd a number. So the only possible result is 1001.

• If you add all the even numbers and substract all the odd numbers, you get: $-1+2-3+4-....-2017+2018=1009$
• from 1009, you can get to any odd number by changing any of the + signs to - or by changing any of the - signs to +
• e.g.: $-1+2-3-4-....-2017+2018=1001$ or: $1009-2*4=1001$ if +4 is changed to -4
• but note that: if you change any of the signs, it is not sufficient to add or substract the chosen number on the right.
• you have to add or substract two times the chosen number on the right. ( $1009-2*4$ and not $1009-1*4$ )
• therefore, even numbers can´t be obtained on the right, since $odd-even=odd$
• thus, 1001 is the only valid solution here.