1 □ 2 □ 3 □ 4 □ 5 □ 6 □ 7 □ 8 □ 9 = 3 3
Seven of the eight " □ "s above are filled with " + ", and the other one with " − ".
Before which integer should the " − " sign be placed to make the equation true?
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Do you think the phrasing is clear? There's a report that they counted the 5th box instead.
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My suggestion:
Seven of the eight " □ "s are filled with " + " and one with " − ". Before which integer should the " − " be placed to make the equation true?
All 9 digits when added makes 4 5 . One needs to swap sign to make it 3 3 , changing the outcome by 4 5 − 3 3 = 1 2 . This will happen with 1 2 / 2 = 6 .
Just wanted to make a de-algebra version, since this problem is solvable by students who haven't had algebra yet.
I did the same way.
To solve this problem, we are given that 7 of the 8 operation signs in this sequence are addition symbols, and one is a subtraction symbol. First, we must add all of these numbers, which yields the result of 4 5 . We then have to realize that 4 5 is 1 2 greater than the number value which we must obtain, which is 3 3 . Secondly, we can see that to get this result, we will have to subtract one of the integers from this sequence, which we will call y . If the value of the set of numbers added together, S is 4 5 , then this will be 4 5 − x . When we subtract x from the sequence however, we are subtracting the number twice, once from the original sequence, and then adding its negative value back into S . In this context, we are concerned with the magnitudes of the numbers, so we can see this through this:
∣ − x ∣ + x = x + x = 2 x
Since the value of all of the numbers deviates 1 2 units from the value of the hidden operations at a value of 3 3 , we can see that 2 x , which is deviation, is equal to 1 2 . Because of this x = 6 . Therefore, the number that we must place the subtraction sign before is 6 .
We need to put minus before 6.
Again, the sum (45)-2x=33 yields the solution.
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Relevant wiki: Arithmetic Puzzles - Fill in the Blanks
We note that 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 4 5 . If we replace an integer k between 1 and 9 with 0, the sum of the 9 integers is 4 5 − k . If we now replace 0 with − k , the sum of the nine integers is 4 5 − 2 k . Therefore, 4 5 − 2 k = 3 3 , we have k = ( 4 5 − 3 3 ) / 2 = 6 .