5
$\begingroup$

Just a quick, easy sum to pass the time for those at work.

This was used as an exercise when I did a programming course to outline the importance of thinking outside the box. It was also used to demonstrate that just because an answer is right, does not mean other answers are wrong.

1 + 1 = 10

How is this so?

Also,

1 + 1 = 11

How is this the case?

$\endgroup$
9
  • $\begingroup$ are both equation with the same system, or two diferent way to add 1 and 1? $\endgroup$
    – Kepotx
    Mar 22, 2018 at 14:35
  • 6
    $\begingroup$ Here is another equation: 1+1=0 in Z/2Z. $\endgroup$ Mar 22, 2018 at 19:45
  • 2
    $\begingroup$ I'd bet one reason the answers came in so quickly is "...when I did a programming course...". I also immediately thought "I bet binary is coming in somewhere". I'd bet it you made the description a little more vague, it would be a little trickier. $\endgroup$
    – BruceWayne
    Mar 22, 2018 at 22:39
  • 5
    $\begingroup$ @BruceWayne Alternatively, a significant portion of SE participants arrive through Stack Overflow and already have programming on the brain anyway. $\endgroup$
    – jpmc26
    Mar 23, 2018 at 1:22
  • 1
    $\begingroup$ There are 10 types of puzzler, those who...(and those who do not) $\endgroup$
    – user41531
    Mar 23, 2018 at 10:22

7 Answers 7

25
$\begingroup$

The first is:

addition of integers in binary.

The second is:

concatenation of strings.

$\endgroup$
6
  • 1
    $\begingroup$ You are indeed correct. There are two more answers to 1 + 1. The obvious one being 2. I cant remember what the fourth answer is though, bonus "points" if you get it! $\endgroup$ Mar 22, 2018 at 14:37
  • $\begingroup$ @JamesDicken A fourth would be 1+1=1 where zero is 0 and any non-zero entity is 1. $\endgroup$ Mar 23, 2018 at 8:06
  • 1
    $\begingroup$ Answering a question like this that is featured on Stack Exchange does wonders for one's reputation... $\endgroup$
    – Statman
    Mar 23, 2018 at 11:12
  • 2
    $\begingroup$ @KlasLindbäck Or perhaps it's boolean logic, where + is logical or (for 1+1=1) $\endgroup$
    – phflack
    Mar 23, 2018 at 13:23
  • 4
    $\begingroup$ 1+1=11 would also work in unary where 1 has been designated as the value. so 1 in this unary is 1 in decimal, and 11 in unary is 2 in decimal and so forth. $\endgroup$
    – Justinw
    Mar 23, 2018 at 13:54
14
$\begingroup$

What a coincidence. I'm a programmer at work right now.

The first equation is correct if

the calculation is performed in base 2 (binary).

The second equation is correct if

interpreted as a string concatenation operation.

$\endgroup$
4
  • 2
    $\begingroup$ Started writing this before Statman's answer but he FGITW'd me. $\endgroup$
    – F1Krazy
    Mar 22, 2018 at 14:37
  • 1
    $\begingroup$ I did, but I fumbled my gun and had to edit my answer after posting, so we'll call it a draw! $\endgroup$
    – Statman
    Mar 22, 2018 at 14:39
  • 2
    $\begingroup$ Unfortunately it will only let me accept one answer which goes to Statman purely based on being the first $\endgroup$ Mar 22, 2018 at 14:52
  • 1
    $\begingroup$ “I'm a programmer at work right now“ — No, you’re not. $\endgroup$
    – Johannes
    Mar 24, 2018 at 15:29
14
$\begingroup$

In the first addition, the two numbers are written is base 2. In the second addition, the two numbers are written in base 1 (and the only digit is 1).

$\endgroup$
6
  • 8
    $\begingroup$ Actually I think both of these answers are correct. $\endgroup$
    – Statman
    Mar 22, 2018 at 14:42
  • 2
    $\begingroup$ Apologies, now I have been further enlightened on the matter I can see they are both correct. However I was referring to string concatenation in the second equation as has been stated by Statman and several others in their answers. $\endgroup$ Mar 22, 2018 at 14:50
  • 2
    $\begingroup$ I disagree. In Base n, the symbols that are used for digits are from 0 thru n-1 e.g., binary or base 2, the digits are either 0 or 1. In base 10, the digits are 0 through 9. In base-16, the digits are [0-9a-f]. In a way, base one is nonsensical. $\endgroup$ Mar 22, 2018 at 22:51
  • 6
    $\begingroup$ Base 1 is often referred to as a "tally system". $\endgroup$
    – CollinD
    Mar 23, 2018 at 0:27
  • 5
    $\begingroup$ @HappyGreenKidNaps: You can disagree all you want. Doesn't make you right: en.wikipedia.org/wiki/Unary_numeral_system $\endgroup$
    – Chris
    Mar 23, 2018 at 14:22
7
$\begingroup$

The first equation:

binary addition, as in base 2, 01+01 = 10

The second equation:

string concatenantion, as "1"+"1" make the string "11"

as you mention in another answer comment, there are other possible equation:

1 + 1 = 2

integer addition

1 + 1 = 11

Gray code, another way to count with 0 and 1

$\endgroup$
3
  • 1
    $\begingroup$ All correct, very well explained as well. Also thank you for filling in the missing gap in my memory. $\endgroup$ Mar 22, 2018 at 14:48
  • 1
    $\begingroup$ @JamesDicken that's because you need 15 reputation to be able to vote. You can learn more about privileges here $\endgroup$
    – Kepotx
    Mar 22, 2018 at 14:52
  • 1
    $\begingroup$ Literally just got the 15 rep as I commented so previous comment has been edited. Thank you for informing me though. $\endgroup$ Mar 22, 2018 at 14:53
4
$\begingroup$

the remaining answer might be:

"b", as summing the codepoints of the two 1 characters (which is how some languages may handle addition of characters) results in 98, the codepoint for b

$\endgroup$
1
$\begingroup$

To potentially address alternate additions as answers have already been posted:

Thinking slightly (way?) outside of the box:

1 + 1 = 1 as in if in PHP 1 represents an array that has a non-numeric key, then the sum of the array with itself is the original array.

Or if the conditions were slightly modified:

1 + 1 = 12 where again in PHP 1 represents an array; however, the key is numeric the addition of the array to itself then would result in a appending the original array to itself with a new key.

$\endgroup$
-2
$\begingroup$

1+1=0 in modular calculus like in (1+1) mod 2

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.