$\large 1 \; \square \; 2 \; \square \; 3\; \square \; 4\; \square \; 5 \; \square \; 6 \; \square \; 7 \; \square \; 8 \; \square \; 9 = 33$

Seven of the eight " $\square$ "s above are filled with " $+$ ", and the other one with " $-$ ".

Before which integer should the " $-$ " sign be placed to make the equation true?

2
3
4
5
6
7
8
9

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Relevant wiki: Arithmetic Puzzles - Fill in the BlanksWe note that $1+2+3+4+5+6+7+8+9 = 45$ . If we replace an integer $k$ between 1 and 9 with 0, the sum of the 9 integers is $45-k$ . If we now replace 0 with $-k$ , the sum of the nine integers is $45-2k$ . Therefore, $45-2k = 33$ , we have $k = (45-33)/2 = \boxed{6}$ .