Can anybody solve this?

Assume there's a way to write
10 10 10 10=7

Then, there's a way to write
10 10 10=7/10 OR 10 10 10=70 OR 10 10 10=-3/10 OR 10 10 10=20.
I'm pretty sure all of these are impossible. But I would love to be proved wrong.

In the operations for any of the '10 10 10 10' set in this solution range of [1,10], there has to be a division operation somewhere, to eliminate a factor of 10, so that really leaves you with three possible operations to take in order to get a value that only needs to be divided by 10.

We can brute force this.
All possible numbers obtainable with +,-,*,/ with two tens:
>100, 1, 20, 0 (1)
All possible numbers obtainable with +,-,*,/ with three tens:
>-90, -10, -9, 0, 0.1, 0.5, 2, 9, 10, 11, 30, 90, 110, 200, 1000 (2)
It is now easy to determine that solutions don't exist for 4, 6, and 7.

Your professor is wrong or you made a mistake in phrasing the problem.

4, 6 and 7 are seemingly impossible to solve

At this point I gotta ask, I cannot come up with the solution for a quad 10 set that is equivalent to 4. Any operations I can think of are either outside of parameters or requires a larger set of 10s.

I would love to see how your Prof. solved for 4, because frankly at this point any insight can only be educational.

Great, thanks user

1+0 1+0 1+0 1+0 = 4

I was thinking about writing a quick script to essentially solve for all possible solutions to a quad 10 set given a +,-,*,/ operation set, and then just search through the results and see if 4,6,7 pop up anywhere.

I may do that, though it is rather late right now and I spent a good hour mentally computing for a set of operations that would give a solution of 4 to a quad 10 set.

Anyway thanks user for that, if OP delivers the solution, I would love to, at this point, be taken to school on how to solve the 4,6,7 ones.

Oh get out of here with that.

My original idea to bullshit the problem is to change the base for the 4 problem.

Essentially I would change that specific quad '10' set from base 10 to base 2, making 10 in base 2 equal to 2 in base 10.

Then I would do 10 + 10 - 10 + 10 = 4 or in base 10 : 2 + 2 - 2 + 2 = 4.

The solution is there, it takes just a glance to see that there is no way to add/multiply/subtract/divide 10 to the numbers (2) in to obtain 4,6, and 7.

did your professor specifically say that you're not allowed to put the math symbols between the 1 and 0 in the numbers 10?

You are adding limitations that don't exist, which is transforming this possible problem into an impossible one

Holy shit, you can use a 10 if you turn it into a zero. And no one said you werent allowed to use zeros.

10*0+10*0+10*0+10*(whateverthefuckyouwant/10)

10+10+10+10=1 in [math]\mathbb{Z}_{39}[/math]
10+10+10+10=2 in [math]\mathbb{Z}_{38}[/math]
10+10+10+10=3 in [math]\mathbb{Z}_{37}[/math]
10+10+10+10=4 in [math]\mathbb{Z}_{36}[/math]
10+10+10+10=5 in [math]\mathbb{Z}_{35}[/math]
10+10+10+10=6 in [math]\mathbb{Z}_{34}[/math]
10+10+10+10=7 in [math]\mathbb{Z}_{33}[/math]
10+10+10+10=8 in [math]\mathbb{Z}_{32}[/math]
10+10+10+10=9 in [math]\mathbb{Z}_{31}[/math]
10+10+10+10=10 in [math]\mathbb{Z}_{30}[/math]

>the answer is whatever the answer is

Yes, this is the true wisdom of the math deity.

bump

4, 6, 7 are impossible

I wrote a quick and dirty python script to try random ones, this is what I got.

0
3 10/(((10)+(10)/10)) 0
4 10/(10)*10/10 1
17 10/10+10/10 2
20 10+(10)*((10-(10))) 10
218 (((10*10-10))/(10)) 9
221 (10-(10+10)/10) 8
888 10*10/(10+10) 5
935 (10+(10)+(10))/(10) 3
100000
200000
300000
400000
500000
600000
700000
800000
900000
1000000
1100000
1200000
1300000
1400000
1500000
1600000
1700000
1800000
1900000
2000000

You see the 8 possible ones were found within 1000 random tries, then I ran 2,000,000 and didn't get any. I also ran this multiple times, no answers. Unless it's using some other operations or something unconventional, they're impossible.

cheating piggot

>random tries
doesn't prove anything

are there really even 2000000 combinations of operations and parenthesis? Can you just do an exhaustive search?

Yeah I know it's not a proof, obviously math doesn't work like that. I'm just too lazy to figure out a way to enumerate all the possible expressions and 2 million probably covers them all.

Brute forced it. Here are all the numbers you can get. 4, 6, 7 are impossible.
-990
-900
-190
-100
-99
-90
-80
-20
-19
-10
-99/10
-19/2
-9
-8
-10/9
-1
-9/10
-1/9
0
1/100
1/20
1/11
1/9
1/5
1/3
9/10
10/11
1
11/10
10/9
2
3
5
8
9
19/2
99/10
10
101/10
21/2
11
12
19
20
21
40
80
90
99
100
101
110
120
190
200
210
300
400
900
990
1010
1100
2000
10000

Brainlet