TOPIC: [Forum Game] 1 + 1 = 11

Hello there.

Here is a new forum game, you must answer the question correctly to win 100k in-game on DarkRP, or you can choose to have 500 points on Deathrun.

Rules

1. You have 2 chances, your first 2 posts are the ones, I will look at.

2. First person to get it, gets it.

3. Winner is decided on Friday this week.

4. I know you will look on the internet, and that is allowed, but good luck finding it.

5. This is maths, BEWARE

6. You can use any math related topic (When I say this, I mean anything with numbers)

7. On Thursday I will give a hint.

Now for the question...


1 + 1 = 11



How is this possible


Oh and by the way, it is VERY unlikely any of you will get it > : )

Have fun!

Disclaimer

The Gamer Guy, is not responsible for any hate towards me or your X box controllers.

But it MIGHT not be possible.

My Maths teacher told my class this (Class of 12 year-olds) And they figured it out in the end.

It's like 1 + 1 = Window. Still have no idea how it works.

So I just placed my 2 toothpicks that each looks like a 1 next to each other.. I got 11 hehehe
1+1= 2 toothpicks ( or any 2 things that is straight and laced next to each other.)
Or the windows method that you and blocked mentioned
1 group of 5 things and 1 group of 6 things
So 1+1= 2 groups of 5 and 6 =11

JohnnySmith wrote:
Not really the answer I was looking for xD, but still a good theory.

This sum shows that the 2 (the sum of 1 and 1) and 11 are congruent. This means there has to be a modulus, which will be (mod n), which is found by finding the positive integer (n) that when 2 or 11 are divided by it, produces the same remainder.

When you divide 2 by any number, the remainder will always be 2 as you can't divide 2 by an integer larger than its self without getting this exact number as a remainder.

So, all you have to find for the modulus is a number that, when you divide it into 7, you get a remainder of 2. So, lets try these solutions:

11/1 = 11
11/2 = 5r1
11/3 = 3r2
11/4 = 2r3
11/5 = 2r1
11/6 = 1r5
11/7 = 1r4
11/8 = 1r3
11/9 = 1r2
11/10 = 1r1
11/11 = 11

Therefore, you have two solutions:

1+1 = 11 MOD3 OR 1+1 = 11 MOD9

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This is what I used, found it by an online answer saying how a sum can equal something that it isn't.
simple.wikipedia.org/wiki/Modular_arithmetic

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I doubt that's what a teacher for 12 year olds would tell them since I don't even know what you mean by MOD

I'm a 12 year old being told about MODS

I have no idea what your talking about, can someone confirm his answer, as it may be a new possible answer, as side from the one I have.

TheGamerGuy wrote:
No one can understand it to confirm it heheh

JohnnySmith wrote:
Lol

Poor Chris and his scientific gamer ways.

My head hurts...

If you really need a hard question, you should have asked:

"Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which."

EDIT: Also the answer is long as hell, but fahk it, I am going to quote it xD

"Does da mean yes if and only if you are False, if and only if B is Random?[1]
Equivalently:

Are an odd number of the following statements true: you are False, da means yes, B is Random?
It was observed by Roberts (2001) and independently by Rabern and Rabern (2008) that the puzzle's solution can be simplified by using certain counterfactuals.[3][4] The key to this solution is that, for any yes/no question Q, asking either True or False the question

If I asked you Q, would you say ja?
results in the answer ja if the truthful answer to Q is yes, and the answer da if the truthful answer to Q is no (Rabern and Rabern (2008) call this result the embedded question lemma). The reason it works can be seen by looking at the eight possible cases.

Assume that ja means yes and da means no.
True is asked and responds with ja. Since he is telling the truth, the truthful answer to Q is ja, which means yes.
True is asked and responds with da. Since he is telling the truth, the truthful answer to Q is da, which means no.
False is asked and responds with ja. Since he is lying, it follows that if you asked him Q, he would instead answer da. He would be lying, so the truthful answer to Q is ja, which means yes.
False is asked and responds with da. Since he is lying, it follows that if you asked him Q, he would in fact answer ja. He would be lying, so the truthful answer to Q is da, which means no.
Assume ja means no and da means yes.
True is asked and responds with ja. Since he is telling the truth, the truthful answer to Q is da, which means yes.
True is asked and responds with da. Since he is telling the truth, the truthful answer to Q is ja, which means no.
False is asked and responds with ja. Since he is lying, it follows that if you asked him Q, he would in fact answer ja. He would be lying, so the truthful answer to Q is da, which means yes.
False is asked and responds with da. Since he is lying, it follows that if you asked him Q, he would instead answer da. He would be lying, so the truthful answer to Q is ja, which means no.
Regardless of whether the asked god is lying or not and regardless of which word means yes and which no, you can determine if the truthful answer to Q is yes or no. If, however, the god is answering randomly, the answer to the Question is still without meaning.

The solution below constructs its three questions using the lemma described above.[3]

Q1: Ask god B, "If I asked you 'Is A Random?', would you say ja?". If B answers ja, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If B answers da, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is not Random. Either way, you know the identity of a god who is not Random.
Q2: Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: "If I asked you 'Are you False?', would you say ja?". Since he is not Random, an answer of da indicates that he is True and an answer of ja indicates that he is False.
Q3: Ask the same god the question: "If I asked you 'Is B Random?', would you say ja?". If the answer is ja, B is Random; if the answer is da, the god you have not yet spoken to is Random. The remaining god can be identified by elimination.
Random's behavior[edit]
Most readers of the puzzle assume that Random will provide completely random answers to any question asked of him; however, Rabern and Rabern (2008) have pointed out that the puzzle does not actually state this.[3] And in fact, Boolos' third clarifying remark explicitly refutes this assumption.

Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
This says that Random randomly acts as a false-teller or a truth-teller, not that Random answers randomly.

A small change to the question above yields a question which will always elicit a meaningful answer from Random. The change is as follows:

If I asked you Q in your current mental state, would you say ja?[3]
This effectively extracts the truth-teller and liar personalities from Random and forces him to be only one of them. By doing so the puzzle becomes completely trivial, that is, truthful answers can be easily obtained.

Ask god A, "If I asked you 'Are you Random?' in your current mental state, would you say ja?"
If A answers ja, A is Random: Ask god B, "If I asked you 'Are you True?', would you say ja?"
If B answers ja, B is True and C is False.
If B answers da, B is False and C is True. In both cases, the puzzle is solved.
If A answers da, A is not Random: Ask god A, "If I asked you 'Are you True?', would you say ja?"
If A answers ja, A is True.
If A answers da, A is False.
Ask god A, "If I asked you 'Is B Random?', would you say ja?"
If A answers ja, B is Random, and C is the opposite of A.
If A answers da, C is Random, and B is the opposite of A."

1+1=2 everyone here is high wow.

Maybe its without the + so it becomes 11?

It depends on what your definition of "is" is.

More people should have a chance xD

BUMP

1+1 is not 11

You know if you put this / into =

I cant do it i dont have the icon on my ipod but that is the anwser i guess

1+1/=11

Good method...

But this should help...

HINT TIME

The answer is related to a famous code used by computers and such.

If that makes any sense.

You all have until tomorrow at 3PM (Great Britain time)

Binary? 1101 0101 1111 1011 Kinda like that I have been studying this, Kinda in school.

Found this:
0 0 0 1 numerical value 20
0 0 1 0 numerical value 21
0 1 0 0 numerical value 22
1 0 0 0 numerical value 23

1111: Being thousands?
1111: Being Hundreds?
1111: Being Tens?
1111: Being Units?

I kinda remember something to do with that.

EDIT: 01001001 00100000 01101100 01101111 01110110 01100101 00100000 01011010 01000001 01010010 01010000 00001101 00001010

That means "I love ZARP" in binary

Just spoke with Chris and realized the gibberish I noticed I'm speaking.

It wouldn't be 1 + 1 = 11, it would be 1+1 = 10

Each binary 'bit' is equivalent to it's 'bit' position from the right hand side as a power of 2 (each value doubles from one to the next) That means like this:

21	20
2	1
1	0

So, 1 + 1 in denary terms does equate to 10 in a base-2 output, if that's what you were meaning, because 11 in binary is equal to 3 (2 + 1).