Add parentheses to make this true

Question

Add parentheses to the following to make a true equation:

$10-9\times8-7\times6-5\times4-3\times2-2\times1=1$

Answer 1

No matter how you place the parentheses, the result will never be 1, because it is always even.

How I found the solution:

Let's take a look at the equation mod 2:
$0+1\times0+1\times0+1\times0+1\times0+0\times1=1$
We start with $0\times1=1$ which is clearly false. Then either $1\times$ or $0+$ is added to the left of the equation. Both of these will never modify the expression to their right, no matter when they are executed.
This means we can continue this chain as often as we like and the result will always stay even.

Answer 2

To take things from a literal computational perspective:

$10-9\times8-7\times6-5\times4-3\times2-2\times(1=1)$
Because $(1=1) == true$ in many languages such as Wolfram Alpha
http://www.wolframalpha.com/input/?i=10-9*8-7*6-5*4-3*2-2*(1%3D1)

Important to note that truthy values can be very lenient:

So $-2 \times true - 130$ still evaluates to $true$ (in fact any value $\ne 0$ would)
So this is "True" in the computational sense rather than the mathematical sense

Answer 3

Since w l's proof doesn't really work, I used a simple python program to iterate through all possible placements, and the answer is:

There is no solution

And the program is a simple recursion:

import itertools

def recurse(part):
    res = set()
    for i in range(1, len(part), 2):
        left = part[:i]
        right = part[i + 1:]
        left = recurse(left) if len(left) > 1 else left
        right = recurse(right) if len(right) > 1 else right
        for x, y in itertools.product(left, right):
            res.add(x - y if part[i] == '-' else x * y)
    return res

1 in recurse([10, '-', 9, '*', 8, '-', 7, '*', 6, '-', 5, '*', 4, '-', 3, '*', 2, '-', 2, '*', 1])

To verify that it works without the typo, you can run:

1 in recurse([10, '-', 9, '*', 8, '-', 7, '*', 6, '-', 5, '*', 4, '-', 3, '*', 2, '-', 1])

Answer 4

That's the best I have so far.

Since there is no base specified, until I find an other solution, I'm going to use the loophole of computing everything in base 11.
$10-(9\times8-7\times6-5\times4-3\times(2-2\times1))_{11}$ =
$11-(9\times8-7\times6-5\times4-3\times(2-2\times1))_{10}$ =
$11 - (72 - 42 - 20 - 0) $ =
$11 - 10 = 1$

Answer 5

I was wondering if the transformation of

$10-9\times8-7\times6-5\times4-3\times2-2\times1 = 1$
into
$10-9\times8-7\times6-5\times4-3\times2-2\times1 - 1 = 0$ is acceptable

Because if so, the answer could be

$1(0(-9\times8-7\times6-5\times4-3\times2-2\times1)(-1)) = 0$

Answer 6

Easy, this is true for

$x=\frac9{142}$, without any parentheses at all.

Explanation:

$10−9\times8−7\times6−5\times4−3\times2−2\times1=1$
is the same as $10 - 9x8 - 7x6-5x4-3x2-2x1 = 1$
which is the same as $10 - 142x = 1$,
which is true for $x = 9/142$. See it on Wolfram Alpha!

Of course,

this only works if you treat the $\times$ sign as an actual $x$, and not as a multiplication...

Answer 7

I found all 16 797 possible solutions, but none of them are equal to 1.

Since pastebin doesn't allowed me to add all answers in one bin, I had to split them in two bins.
The first 8000 - https://pastebin.com/YE0FqQpm
The rest 8797 - https://pastebin.com/GuVvv1QF

Answer 8

I'll give another solution, related to the answer given by Marcus

Using base 15.
$(10-(9\times8-7\times6-(5\times4-3\times2)-2\times1))_{15}$ =
$(10-(4\text{C}-2\text{C}-(15-6)-2))_{15}$ =
$(10-(20-\text{E}-2))_{15}$ =
$(10-(11 - 2))_{15}$ =
$(10-\text{E})_{15}$ =
$1_{15}$

Answer 9

1 (0-9×8-7×6-5×4-3×(2-2))×1=1

Never specified whether I can break up numbers

Answer 10

After focusing a while, I think I have found the answer.
According to me the parenthesis would be added in following sequence:

(10-9) * (8-7) * (6-5) * (4-3) * (2-1)=1
As
1) 10-9=1:- (1) * (8-7) * (6-5) * (4-3) * (2-1)=1
2) 8-7=1:- (1) * (1) * (6-5) * (4-3) * (2-1)=1
3) 6-5=1:- (1) * (1) * (1) * (4-3) * (2-1)=1
4) 4-3=1:- (1) * (1) * (1) * (1) * (2-1)=1
5) 2-1=1:- (1) * (1) * (1) * (1) * (1)=1

Hence,

1 * 1 * 1 * 1 * 1=1

Answer 11

It would look something like the following (see explanation because the last couple of steps depend on pen and paper)

([(10-9*8-7*6-5*4-3)*(2-2)]) * (1)=1

Reasoning:

1) Put a parenthesis around (2-2) so it becomes zero and zeros out everything to it's left.

2) So you have this without the final pairs of parenthesis:
0*1=1

3) Draw the final pairs of parentheses so they intersect with the multiplication sign's bottom four points, leaving a straight line
(0)*(1) = 1

4) Which, without the parentheses, is
0 | 1 = 1

5) Since "|" is the bitwise OR and 1 normally represents TRUE, this is a true expression