I can't believe there are adults who can't do this
(media.scored.co)
They clearly forgot the order of operations.
New here?
Create an account to submit posts, participate in discussions and chat with people.
You must log in or sign up to comment
25 comments:
Do everything with a defined number after the "=" sign first.
https://communities.win/c/ConsumeProduct/p/199OdIIyP2/
How did I do
How did I do
Plug 4 in the left column
Follow the order of operations
Well, that much worked, so we're dealing with idiots.
Follow the order of operations
Well, that much worked, so we're dealing with idiots.
To be quite honest, this is basically sodoku with simple math equations. Why should this be involved in education?
You can remake it easily in Excel or with pen/paper. I am conscidering giving something like this to my kids to earn another piece of Halloween candy.
It's like a simple form of Sudoku. You always do the line in which only 1 is missing.
1. Column: 9 - 4 / 8 * x = 7 --> x is 4.
4. Row: x + 1 + 7 - 4 = 8 -> x is 4. It's the same x as above.
3. Column: 6 - 5 + y - 7 = 3 -> y is 9.
3. Row: 8 - 5 = y - z * 2 -> y is 9 as above, therefore z is 3.
From here on you have 2 or 3 missing in every line. That means we have 5 equations each with 2 or 3 variables (6 in total). We need to figure out a viable constellation of values so that all 5 equations are correct. So first the ones with 2 missing (plus a simplified form for later):
I 1. Row: 9 + a - 6 = b - 3 | b = a + 6 | a = b - 6
II 2. Row: 4 + c * 5 - d = 6 | d = 5c - 2 | c = d/5 + 0.4
III 5. Row: 7 = e * 3 - f - 2 | f = 3e - 9 | e = 3 + f/3
IV 2. Column: a = c + 5 - 1 - e | a = c - e + 4 | e = c - a + 4 | c = a + e - 4
V 4. Column: b = d - 3 - 4 + f | b = d + f - 7 | d = b - f + 7 | f = b - d + 7
So we have to combine equations. In theory there can be multiple solutions, but we are restricted to integers from 1 to 9. Each variable must be one of those.
I and IV: c - e + 4 = b - 6 -> b = c - e + 10 | e = c - b + 10
I and V: a + 6 = d + f - 7 -> a = d + f - 13
II and V: 5c - 2 = b - f + 7 -> b = f + 5c - 9
II and IV: d/5 + 0.4 = a + e - 4 -> a = d/5 - e + 4.4 | e = d/5 - a + 4.4 | d = 5e + 5a - 22
III and IV: 3 + f/3 = c - a + 4 -> a = c - f/3 + 1
III and V: 3e - 9 = b - d + 7 -> b = 3e + d - 16
So what did we get? We managed to mix up variables. But otherwise nothing really. But if we get just 1 variable, we'd get all others too. The problem is that we can't. So we'll need to trial and error. But, we have a lot of options, and I assume some multi-dimensional math or drawing would be required. I could write a program to do it, or I'd have to spend time to figure out a way to meaningfully draw it to a paper.
But this is way too hard (almost impossible) for fifth grade. At that time I was by far the best in class in Math, and even I would have struggled with it. Maybe with trial and error, and with that 1-9 hint, but to do it properly takes quite some time and skills that are... possibly even inaccessible to 95%+ of people (non-humans not counted). And for the remaining ones it would take quite some time. ~1% or less would be able to do it without significant time.
1. Column: 9 - 4 / 8 * x = 7 --> x is 4.
4. Row: x + 1 + 7 - 4 = 8 -> x is 4. It's the same x as above.
3. Column: 6 - 5 + y - 7 = 3 -> y is 9.
3. Row: 8 - 5 = y - z * 2 -> y is 9 as above, therefore z is 3.
From here on you have 2 or 3 missing in every line. That means we have 5 equations each with 2 or 3 variables (6 in total). We need to figure out a viable constellation of values so that all 5 equations are correct. So first the ones with 2 missing (plus a simplified form for later):
I 1. Row: 9 + a - 6 = b - 3 | b = a + 6 | a = b - 6
II 2. Row: 4 + c * 5 - d = 6 | d = 5c - 2 | c = d/5 + 0.4
III 5. Row: 7 = e * 3 - f - 2 | f = 3e - 9 | e = 3 + f/3
IV 2. Column: a = c + 5 - 1 - e | a = c - e + 4 | e = c - a + 4 | c = a + e - 4
V 4. Column: b = d - 3 - 4 + f | b = d + f - 7 | d = b - f + 7 | f = b - d + 7
So we have to combine equations. In theory there can be multiple solutions, but we are restricted to integers from 1 to 9. Each variable must be one of those.
I and IV: c - e + 4 = b - 6 -> b = c - e + 10 | e = c - b + 10
I and V: a + 6 = d + f - 7 -> a = d + f - 13
II and V: 5c - 2 = b - f + 7 -> b = f + 5c - 9
II and IV: d/5 + 0.4 = a + e - 4 -> a = d/5 - e + 4.4 | e = d/5 - a + 4.4 | d = 5e + 5a - 22
III and IV: 3 + f/3 = c - a + 4 -> a = c - f/3 + 1
III and V: 3e - 9 = b - d + 7 -> b = 3e + d - 16
So what did we get? We managed to mix up variables. But otherwise nothing really. But if we get just 1 variable, we'd get all others too. The problem is that we can't. So we'll need to trial and error. But, we have a lot of options, and I assume some multi-dimensional math or drawing would be required. I could write a program to do it, or I'd have to spend time to figure out a way to meaningfully draw it to a paper.
But this is way too hard (almost impossible) for fifth grade. At that time I was by far the best in class in Math, and even I would have struggled with it. Maybe with trial and error, and with that 1-9 hint, but to do it properly takes quite some time and skills that are... possibly even inaccessible to 95%+ of people (non-humans not counted). And for the remaining ones it would take quite some time. ~1% or less would be able to do it without significant time.
Toast message
1 7
2 8
9 3
4
5 6? (from top left to bottom right)
Also, this is not a standard 5th grade math assignment. To do it purely with logic and no trial and error takes some thought, particularly when getting to the third column with 1, 5, 6, and 7 left over if done as easy as possible.