4-2^2+5x3-2=11

[thegreenhell]

FredClogs said:

This is a beautiful example, I do believe you're the only one who read this and thought to put brackets after the equals. Bravo.

And possibly the only one who's read 4-2 and got -2

[FredClogs]

thegreenhell said:

FredClogs said:

This is a beautiful example, I do believe you're the only one who read this and thought to put brackets after the equals. Bravo.

And possibly the only one who's read 4-2 and got -2

Goddam this fking question

[4x4Tyke]

El Joffo said:

By jove you're right. Ruins that idea!

Not entirely, I never even considered putting brackets on the right hand side.

[stuthemong]

El Joffo said:

........... = 1(1)

This was genius. Robbed! Deserved more

[jet_noise]

stuthe said:

El Joffo said:

........... = 1(1)

This was genius. Robbed! Deserved more

That assumes that implicit operators are acceptable/available doesn't it?
i.e. 1(1)= 1x(1)= 1x1

[Xaero]

4-2^2+5x3-2=11
(4-2)^2+5x3-2=11
(2)^2+5x3-2=11
4+5x3-2=11
4+(5x3)-2=11
4+(15)-2=11
19-2=11
17=11

4-2^2+5x3-2=11
4-(2)^2+5x3-2=11
4-4+5x3-2=11
4-4+(5x3)-2=11
4-4+15-2=11
15-2=11
13=11

4-2^2+5x3-2=11
((4)(-2)^2)+5x3-2=11
(4*4)+5x3-2=11
16+5x3-2=11
16+15-2=11
29=11

4-2^2+5x3-2=11
4-2^(2+5)x3-2=11
4-2^7x3-2=11
4-64x3-2=11
-60*3-2=11
-180-2=11
-182=11

4-2^2+5x3-2=11
4(-2)^2+5x3-2=11
4+4+5x3-2=11
4+(4+5)x3-2=11
4+9x3-2=11
4+9x1=11
4+9=11
13=11

4-2^2+5x3-2=11
(4-2)^2+5x3-2=11
2^2+5x3-2=11
4+5x3-2=11

fk it.

[wilwak]

jet_noise said:

stuthe said:

El Joffo said:

........... = 1(1)

This was genius. Robbed! Deserved more

That assumes that implicit operators are acceptable/available doesn't it?
i.e. 1(1)= 1x(1)= 1x1

The fact that the original equation includes a ‘x’ for multiplication makes the use of an implied multiplication in the same equation a bit naughty.

I still reckon it’s a misprint.

[Zod]

Einion Yrth said:

El Joffo said:

-2^2 can be either -4 or +4 depending on the context.

Edited by El Joffo on Thursday 11th October 18:56

No it can't; roots have positive and negative solutions, squares are always positive.

What about i^2?

[annodomini2]

Is the right hand number hexadecimal? i.e. 17?

[dodgyviper]

Methinks typo

For this to work, as the final answer is a prime we would look to make it by
:- multiplying 1 and 11 by building up the brackets to include all Minuses and Additions before multiplication to make 1 * 11.
:- Or we have standalone minus/addition(s) that occur(s) after multiplication, thereby allowing us to use numbers other than 1 and 11

Scenarios
(A) both minuses are standalone,
(B) only the first minus is standalone
(C) only the second minus is standalone
(D) only the addition is standalone
(E) First minus and addition are standalone
(F) Second minus and addition are standalone
(G) Both minuses and the addition are standalone
(H) Addition and subtraction occurs before all multiplication (i.e. they are nested in brackets)

For (A) This leads to 4 - 2^2+5*3 -2 where we need to get 2^2+5*3 to calculate out as -9
Not going to happen, can't multiply positives to get a minus no matter how the brackets lie

For (B) This leads to 4 - 2^2+5*3-2 where we need to get 2^2+5*3-2 to calculate out as -7
Not going to happen, can't multiply anything other than -1 and 7 (or 1 and -7) to get -7 (prime)
We can get the 1 with the 3-2, but we need to then get -7 from 2^2+5
Options are
(2^2)+5=9
2^(2+5)=128
Both incorrect

For (C) This leads to 4 - 2^2 + 5 * 3 -2 where we need to get 4 - 2^2 + 5 * 3 to calculate out as 13 Not going to happen as 13 is prime
In this scenario, the first minus and the addition must be included within brackets prior to being multiplied, and both products would have to become one of 1 or 13 eg
Options are:
(4-2)^(2+(5*3))
(4-2)^((2+5)*3)
4(-2^((2+5)*3))
4(-2^(2+(5*3))

4 multiplied by any whole number in brackets will never get to 13 or 1, so the last two can be discarded
4-2=2 so can be discarded
2+(5*3)=17 so can be discarded
(2+5)*3=21 so can be discarded
-2^(2+(5*3))= -131,072 so can be discarded
-2^((2+5)*3)= -2,097,152 so can be discarded

For (D) This leads to 4 - 2^2 + 5 * 3 - 2.
Options for 4 - 2^2 are
4-(2^2)= 0
(4-2)^2= 4
4(-2^2)= -16
Options for 5 * 3 - 2 are
5*(3-2)= 5
(5*3)-2= 13
None of these add together to make 11

For (E) This leads to 4 - 2^2 + 5 * 3 - 2 The 2^2 must equal 4, so is cancelled out in 4-2^2, so we need to get 5*3-2 to calculate out as 11
only two answers are
5*(3-2)= 5
(5*3)-2= 13
Both are incorrect

For (F) This leads to 4 - 2^2 + 5 * 3 - 2 This forces the 5*3 to be 15, so we have (4 - 2^2) + 15 - 2 therefore 4 - 2^2 needs to come out as -2
Options for 4 - 2^2 are
4-(2^2)= 0
(4-2)^2= 4
4(-2^2)= -16
All are incorrect

For (G) This leads to 4 - 2^2 + 5 x 3-2 where 2^2 can only be 4 and 5*3 can only be 15 - gives final answer of 13 - incorrect

For (H) our options are;
(4 - 2)^((2 + 5 x (3-2)) = 128
(4 - 2)^((2 + 5) x (3-2)) = 128
(4 - 2)^(2 + (5 x 3)-2) = 32.768
4(( - 2^(2 + 5))) x (3-2) = -512
4( - 2^(2 +(5 x 3)-2) = -131,072
4( - 2^(((2 +5 )x 3)-2)) = 2,097,152

Will be more permutations, but multiplying anything by 4 won't make a prime - so the last 3 above are automatically out and the first two are simply powers of 2 - never going to be 11

If it is -11 as an answer then as already noted by others, 4(-2^2) + 5*(3-2) = -11

[Einion Yrth]

Zod said:

Einion Yrth said:

El Joffo said:

-2^2 can be either -4 or +4 depending on the context.

Edited by El Joffo on Thursday 11th October 18:56

No it can't; roots have positive and negative solutions, squares are always positive.

What about i^2?

I'll give you that, but somehow I don't think that a knowledge of complex numbers is supposed to be necessary for the solution of the original problem.

[soupdragon1]

I think this thread is like one of those gifs that you're waiting and waiting for something to happen, and you've sat there for ages until you finally realise you've been had.

[chemistry]

soupdragon1 said:

I think this thread is like one of those gifs that you're waiting and waiting for something to happen, and you've sat there for ages until you finally realise you've been had.

I promise it's all real; the question photo was taken directly from my son's homework, to exclude the possibility of me inadvertently introducing a typo. For completeness, the full question associated with the equation is simply: "Add the necessary brackets to make this equation true."

The bad news is that although he had maths today, the answer wasn't forthcoming because (unsurprisingly) nobody had yet submitted a correct answer. The good news is that he's provided a hint; that brackets are involved on both sides of the equation (so presumably something like (1)(1), as has been suggested earlier). No mention/admission of a typo.

I'll report back next week, once the answer is finally revealed. I'm as keen as any of you to know what the flippin' answer is...I've spent hours on this by now!

[chemistry]

Here’s the original question; note there’s a bag of sweets resting on this...

[thegreenhell]

This is so frustrating. I'm convinced there's a typo making a sign wrong somewhere. I can solve it if the equation begins with -4 or ends with +2, or if the answer is -11.

Alternatively, I'm hoping that it's printed as intended, but either the teacher is a fool or there's been some new method of teaching maths introduced in the last 30 years that manages to reverse polarity in some previously unimaginable way.

[Bill]

In for the disappointment when it turns out to be a typo. Or the teacher simply can't add up...

[hab1966]

If it is -11 as an answer then as already noted by others, 4(-2^2) + 5*(3-2) = -11

Surely that's 21 not -11?

[soupdragon1]

(11) in excel can mean minus 11 with conditional formatting but I don't think that's the path it's going...

[JuanCarlosFandango]

If you can get the first part to -2 you're away. But I can't. No sweeties for me.

[4x4Tyke]

chemistry said:

Here’s the original question; note there’s a bag of sweets resting on this...

What age group/key stage is this supposed to be ?

From the clue, I think we can also conclude we are looking for a value of 1 from 1(1) or (1)(1) because (11) doesn't change the value so there is no point to adding brackets (11) to the RHS.

Given the way the power is written as a superscript, I think that we can conclude that is 'as is' unchanged by brackets.

Given A = "4 - 2²"
The largest value for A is zero when used as is, the smallest is -16 with brackets added.

Given B = "5 * 3 - 2"
Then B will always be positive, the smallest value is 5, from "5 * (3 - 2)" which larger than A.

Therefore A + B = 1 or 11

So it is also actually impossible if the RHS = 1 since we would need a way to get a negative number from B, but we can't.

The clue doesn't actually help, I'm certain there is a transcription or maths error from the teacher.

Edited by 4x4Tyke on Saturday 13th October 08:53