Blank Space Part 7

Logic Level 2

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

There are $2^7 =128$ ways in which we can fill the squares with $+, -$ .

How many ways would make the equation true?

Note : You are not allowed to use parenthesis.

0 16 8 2 4

11 solutions

Halum Singh
Jul 30, 2015

Adding all the numbers we get 36 which is even.

Now we know that, sum and difference of two numbers have same parity i.e. both even or both odd. So changing sign we never get odd result.

That is a hell of a lot smarter way than I figured it out by guessing and checking about 20 and then assuming there was no way to do it.....Good on ya.

Jace Beeny - 5 years, 10 months ago

I did it the hard way too. I disregarded +/- 8 and instead figured I needed to get 1 or 17 for the previous operations. I clustered {1.2,3} and the operations in between and that yielded {-4,0,2,6} and then I clustered {4, 5} and the operations before each, and that yielded {-9,-1,1,9} and finally clustered {6,7} and the operations before each, and that yielded {-13,-1,1,13}
so adding one from each bolded group, I needed to get either 1 or 17. It can't be done. It simplified the 128 combinations, but clearly not as simple as the obvious taking the parity of numbers rule.

Ken Hodson - 5 years, 6 months ago

I'm only commenting on this because it's the first solution - but if we are only allowed to use + and -, is there any way parentheses could make a difference?

Katherine barker - 3 years, 10 months ago

I’m not sure. I don’t think so

Westy Raun - 2 years, 1 month ago

Owen Berendes
Jan 27, 2016

Parity: Since there are 4 odd terms and 4 even terms the result of addition or Subtraction is always even. Therefore the result can never be 9.

Correct! Thank you.

Chung Kevin - 5 years, 4 months ago

Why does there necessarily have to be four of each?

John Bingamon - 3 years, 2 months ago

On the LHS, there are 4 odd number (1, 3, 5, 7) and 4 even numbers (2, 4, 6, 8).

Given that, if all we are allowed to do is add or subtract, then the RHS can only be even, and hence is never odd.

Chung Kevin - 3 years, 2 months ago

can never be odd

Laszlo Kocsis - 3 years, 1 month ago

@Laszlo Kocsis – Oh yes! Thanks for pointing that out. I have edited my comment.

Chung Kevin - 3 years, 1 month ago

@Chung Kevin – The fact that there were 4 odd and 4 even numbers was irrelevant. If it were relevant then one ought to be able to replace the ?'s with +'s and -'s in the following to create an equation, but one cannot. 1 ? 3 ? 5 ? 7 ? 9 ? 11 ? 13 ? 15 = 17.

Steve Brian - 2 years, 8 months ago

@Steve Brian – I'm not sure what you are getting at.

The generalization of the statement is that "If there is an even (resp. odd) number of odd integers on the LHS, then the sum/difference is even (resp. odd)."

Thus, in your case, with 8 odd numbers, the sum/difference must be even. Hence, we can conclude that the RHS must be odd, so there are no solutions to a sum of 17.

Chung Kevin - 2 years, 8 months ago

Vô Người
Aug 8, 2015

Assume we give all '+' at first, so SUM from 1->8 is equal to 36. Now if we change a '+' to '-' before number x then SUM will decrease by 2 x, hence we got 36-2 c=9 with c equal to sum of all number x we put '-' before. This does not seem to be right when 36-9=27 is an odd number , so there is no way to make this equation become true.

I thought the same and generalized it Elegant solution

Bijay Shah - 3 years, 3 months ago

Zakir Dakua
Dec 2, 2015

sum - 9 = 36 - 9 = 27

You can't divide it into two equal integer, so can't make their difference 0. So no way there.

Yes. This is due to the Parity Argument . Thank you for your solution!

Chung Kevin - 5 years, 6 months ago

Dani Paunov
Dec 3, 2017

When you switch the sign of a number, the result changes by that number times two, and since adding all the numbers results in an even number, you cannot make it odd.

Santiago Arce
Nov 11, 2017

The sum of all those numbers is 36. You have to notice that to subtract one of those numbers instead of having added it is like subtracting it twice. So you see if there is any combination of the doubled numbers from 1 to 8 that makes 27 ( which is what needs to be subtracted from 36 to get 9) and you find there are none because they are all even and 27 is an odd number.

Gordon Glenn
Aug 19, 2017

The sum of the first 8 positive integers is 36. Looking for two complementary subsets (of the available set of 8 numbers) that will yield the sum of 36 and a difference of 9, I wrote this linear algebraic system:

x + y = 36

x - y = 9

Using your favorite method to solve this system, you should get x = 22.5 and y = 13.5 Clearly, these rational numbers cannot be made by addition of integers, so the system has no solutions, and therefore there are 0 ways to make the original equation true.

Stacy Blatt
Aug 5, 2017

This simplifies to 36-2x=9, because any number we make negative, we subtract from 36 (all positive) twice; x=12.5, so it can't be done.

A Former Brilliant Member
Jul 19, 2017

We know there must be some additions and subtractions: Let's start with adding all the digits 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 Now, when if we subtract any digit that would mean we need to take away twice its value: ie. if we subtract 6 instead of adding: 1+2+3+4+5-6+7+8 = 36-2*6 = 24

Same holds for multiple digits as well, therefore 1 2 3 4 5 6 7_8 = 36 - 2 x, where x = sum of any number of digits between 1 and 8.

So, if 1 2 3 4 5 6 7_8 = 9 where _ can be + or - Then 36-2x = 9, where x as explained above, sum of digits to be subtracted Which means x = 27/2 or 13.5

A sum of whole numbers can't be a decimal number. Therefore, there will be zero ways in which we can make the equation true.

Adam W-M
Nov 27, 2016

In order for this to work, you must have an odd amount of the number nine, resulting in an odd answer. Essentially, If you alternated adding and subtracting sets of 9 (1+8), (-2,-7) You end up with zero, so there must be 0 solutions

Uddeshya Singh
Sep 10, 2016

addition or subtraction .always even. therefore 9 just can't be formed

Blank Space Part 7

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