7
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One of the numbers here could be changed to a specific value so that every statement is true.

  • One plus one equals two.

  • Two plus two equals four.

  • Five plus ten equals fifteen.

Which number could be changed?   To what possible values?

To change a number means to change the numerical value of a single instance, and not just its spelling, description or interpretation.

Intended solutions are few enough, unambiguous enough and self-explanatory enough so that their results should be completely displayed but need not be explained.

In the spirit of , unintended solutions deserve recognition and are more than welcome if explained as well as demonstrated, but are not considered complete by themselves.

“... a specific value ...” is deliberately worded to limit possible answers.

$\color{black}{\small\sf Hint{\scriptsize\bf\kern.5mu\raise1mu/}warning\!:}$ This is puzzle is more self-referential that it might seem, even after recognizing that it seems self-referential.

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  • $\begingroup$ When you say "a single instance" do you mean that every statement will still be true if exactly one specific value in the bullet points is changed? Or do you mean changing all instances of one specific number to something else? $\endgroup$ – John Clifford Oct 14 '16 at 9:38
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    $\begingroup$ That's what I thought, and I specifically mentioned bullets to see if you'd confirm my suspicion that the number needing changed isn't in them. :) Thanks. $\endgroup$ – John Clifford Oct 14 '16 at 9:49
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    $\begingroup$ A lifetime of Sierra adventure games has prepared me for all the lateral thinking I need. :D $\endgroup$ – John Clifford Oct 14 '16 at 9:51
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    $\begingroup$ I upvoted to change the number between the two arrows, congratulating myself on true lateral-thinking... then I saw "specific value". Damn. $\endgroup$ – Xenocacia Oct 14 '16 at 9:54
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    $\begingroup$ @humn I added the cases with replacing single digits with appropriate notes. $\endgroup$ – oleslaw Oct 18 '16 at 10:09

11 Answers 11

23
+50
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The final (I believe) answer is:

0, 4, 7 or 10, all to replace "One" in the very first statement.

Reasoning:

The one number to change in the original riddle is definitely the "one" in the very first sentence. This value has to be replaced by 0 at some point to stop the recursion. However you could have more than 1 step to achieve this.

CASE#1

0 - The solution is obvious

CASE#2

4 - Then, the riddle becomes:
4 of the numbers could be changed...
1+1=2
2+2=4
5+10=15
The 4 numbers we change are one of the equiations and the first "One". The number we change them to is again 0. Finally we get 0+0=0, which is true and 0 numbers to change.

CASE#3

7 - Then, the riddle becomes:
7 of the numbers could be changed...
1+1=2
2+2=4
5+10=15
The 6 numbers we change are two of the equations and the first "One". The number we change them to is again 0. Finally we get 0+0=0 and 0+0=0, which is true and 0 numbers to change.

CASE#4

10 - Then, the riddle becomes:
10 of the numbers could be changed...
1+1=2
2+2=4
5+10=15
The 10 numbers we change are all the numbers in the riddle. The number we change them to is again 0. Finally we get 0+0=0 three times and 0 numbers to change, which is true.

Additionally if the riddle was written in digits and the change of a number could be a change of a digit (one of the two in two-digit number) there are additional solutions possible (cases #5-#12):
The original riddle would look like:

1 of the numbers here could be changed...
1 + 1 = 2
2 + 2 = 4
5 + 10 = 15

The additional solutions are:

2 (infinite recursion), 3, 5 (leading to 2), 6, 8 (only if we count "00" as "0"), 9, 11 (only if we count "00" as "0" and we are still able to replace whole two-digit number as one) and 12 (only if we count "00" as "0")

CASE#5:

2 - Then the riddle becomes:
2 of the numbers here could be changed...
1 + 1 = 2
2 + 2 = 4
5 + 10 = 15
The two digits we change are 5+10=15 in the last equation.
The number to which we change it is any number from 0 to 9 (not 5), so we get i.e. 0+10=10 which is true and 2 numbers to change.
Then the 2 numbers to change are again the same numbers (0+10=10 in the last equation). Now we change them to another value from 1 to 9. So we get i.e. 5+10=15 (looks familiar, doesn't it?) and still 2 numbers to change.
This way we get a sweet infinitely recursive set of "valid" solutions.

CASE#6

3 - Then the riddle becomes:
3 of the numbers here could be changed...
1 + 1 = 2
2 + 2 = 4
5 + 10 = 15
The 3 digits we change are 5+10=15 in the last equation and the first "One".
The number to which we change it is 0, so we get 0+10=10 which is true and 0 numbers to change. Done.

CASE#7

5 - Then the riddle becomes:
5 of the numbers here could be changed...
1 + 1 = 2
2 + 2 = 4
5 + 10 = 15
The 5 digits we change are 5+10=15 in the last equation and the first "One".
The number to which we change it is 2, so we get 2+20=22 which is true and 2 numbers to change. Doesn't it look like CASE#5 already? :)

CASE#8

6 - Then the riddle becomes:
6 of the numbers here could be changed...
1 + 1 = 2
2 + 2 = 4
5 + 10 = 15
The 6 digits we change are 5+10=15 in the last equation, whole first (or second) equation and the first "One".
The number to which we change it is 0, so we get 0+0=0, 0+10=10 which is true and 0 numbers to change. Done.

CASE#9

NOTE: This case works only if we agree that "00" is a number and its value is actually "0". 8 - Then the riddle becomes:
8 of the numbers here could be changed...
1 + 1 = 2
2 + 2 = 4
5 + 10 = 15
The 8 digits we change are 5+10=15 in the last equation, whole first (or second) equation and the first "One".
The number to which we change it is 0, so we get 0+0=0, 0+00=00 which is true and 0 numbers to change. Done.

CASE#10

9 - Then the riddle becomes:
9 of the numbers here could be changed...
1 + 1 = 2
2 + 2 = 4
5 + 10 = 15
The 9 digits we change are 5+10=15 in the last equation, whole first and second equation and the first "One".
The number to which we change it is 0, so we get 0+0=0, 0+0=0, 0+10=10 which is true and 0 numbers to change. Done.

CASE#11

NOTE: This case works only if we agree that "00" is a number and its value is actually "0". 11 - Then the riddle becomes:
11 of the numbers here could be changed...
1 + 1 = 2
2 + 2 = 4
5 + 10 = 15
The 11 digits we change are 5+10=15 in the last equation, whole first and second equation and the first "One" considered as one number (even if it has 2 digits "11").
The number to which we change it is 0, so we get 0+0=0, 0+0=0, 0+00=00 which is true and 0 numbers to change. Done.

CASE#12

NOTE: This case is very similar to CASE#11. The only difference is to treat the "11" as two digits. 12 - Then the riddle becomes:
12 of the numbers here could be changed...
1 + 1 = 2
2 + 2 = 4
5 + 10 = 15
The 12 digits we change are all the digits in the riddle.
The number to which we change it is 0, so we get 0+0=0, 0+0=0, 0+00=00 which is true and 00 numbers to change. Done.



To understand recursion you must first understand recursion...

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  • 2
    $\begingroup$ You're likely to be the first to circle in on the most technically correct answer, which includes fewer possibilities, though I wouldn't mind being proven wrong. $\endgroup$ – humn Oct 14 '16 at 9:54
  • $\begingroup$ @humn, I believe the most technically correct answer, includes the same number of possibilities! $\endgroup$ – Arth Oct 14 '16 at 16:28
  • $\begingroup$ I tried to think laterally more than I should. Its more recursive than it looked like in the beginning. $\endgroup$ – oleslaw Oct 16 '16 at 17:49
  • $\begingroup$ Great recursive conclusion: To understand recursion you must first understand recursion...! Your endurance will be rewarded by more than upvotes. $\endgroup$ – humn Oct 17 '16 at 6:09
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    $\begingroup$ @GentlePurpleRain The replacement has to be done with a "specific value" - that is replace both numbers with the same value. 7 and 8 are different values. $\endgroup$ – oleslaw Oct 18 '16 at 18:30
9
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5(five) can be changed to 0(zero)

making

One plus one equals two. ( 1 + 1 = 2)

Two plus two equals four. (2 + 2 =4)

and

Five plus ten equals fifteen (5 + 10 = 15)

to

Zero plus ten equals ten. (0 + 10 = 10)

Which is also true.

So the one number in the statements that can be changed and still keep all the statements true is, changing

Five to Zero

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    $\begingroup$ ^vote with a note: This contains the last intended piece of the puzzle, considering together all the answers so far, but still counts as two changes, not one. $\endgroup$ – humn Oct 14 '16 at 10:29
  • $\begingroup$ Nonetheless that last piece isn't being used in the intended way, but I couldn't count this way wrong with a clear conscience if it were put in a different context (you'll have to figure out what that means) $\endgroup$ – humn Oct 14 '16 at 10:40
5
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Change the first sentence:

"One of the numbers here could be changed to a specific value so that every statement is false." (true and false are 1 and 0 in binary)

Then interpret the statements differently:

1 + 1 = 11 (not 2)
2 + 2 = 22 (not 4)
5 + 10 = 510 (not 15)

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  • $\begingroup$ ^vote for lateral thinking. This is the kind of unintended answer I would have to accept as a part of a complete answer. $\endgroup$ – humn Oct 14 '16 at 17:04
4
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Shamelessly peeking at the answers already given,

you could take the "One of the numbers here could be changed to a specific value so that every statement is true" rule

and change it to

9 of the numbers here could be changed to a specific value so that every statement is true

at which point the specific value would be

0

because

0 + 0 = 0 and 0 + 0 = 0 and 0 + 0 = 0

But I doubt that this is in keeping with the spirit of this puzzle because of the two steps...?

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  • $\begingroup$ This is exactly in the spirit of the puzzle but incomplete. $\endgroup$ – humn Oct 14 '16 at 17:00
  • $\begingroup$ For example, is the statement with the changed number still true? $\endgroup$ – humn Oct 14 '16 at 19:16
  • $\begingroup$ So you want this to end in an infinite loop? Sorry but the programmer in me won't let that heppen :P $\endgroup$ – Brent Hackers Oct 15 '16 at 10:46
  • $\begingroup$ (snicker) This language (English#) has a conditional operator not found in most formal ones: could $\endgroup$ – humn Oct 15 '16 at 10:51
3
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I think this might be the answer:

The number that can be changed is the one in "One of the numbers here could be changed to a specific value so that every statement is true."

The possible values are Two or Three.

This would turn it into "Two/Three of the numbers here could be changed to a specific value so that every statement is true."

If one were to then solve the puzzle that is created by that change, you could for example change Two and Two in the second one to One and Three, or change any three numbers to whatever you want.

TL;DR: I suspect you weren't asking us to make a change that still makes the bullet points true, but a change that allows the first statement to be a solveable puzzle. Though in all honesty I'm expecting to be wrong with this as it seems too simple a solution and I get the feeling the number needs to be changed to something that still makes the bullet points true.

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  • $\begingroup$ ^vote with a note: you're on the right track but "a specific number" was deliberately stated to disallow changing to different numbers at the same time. (I understand what you mean by "Two/Three," though, and have no objection to that, but think the ultimate answer is a different variation of it.) $\endgroup$ – humn Oct 14 '16 at 10:17
  • $\begingroup$ Hmm, interesting. You get an upvote for making me think, and I shall continue to ruminate upon the problem. $\endgroup$ – John Clifford Oct 14 '16 at 10:20
3
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To change a number means to change the numerical value of a single instance, and not just the spelling, description or interpretation.

Numbers: One, Two, Four, Five, Ten, Fifteen
Numerical Values: 1, 2, 4, 5, 10, 15

Which number could be changed?   To what possible values?

Any number could be changed. To their respective numerical values.

And Truth will still stand.

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  • $\begingroup$ ^vote with a note: Good lateral thinking but this doesn't really count as a change because the numerical value didn't change. $\endgroup$ – humn Oct 14 '16 at 17:02
  • $\begingroup$ @humn, i have a sinking feeling that the mistake we are making is thinking of 'numerical value' only in a restricted set (0 to 9). Think about this 'One of the numbers X could be changed to a specific value so that every statement is true.' Without using X, our bullet points are already true. $\endgroup$ – WeShall Oct 15 '16 at 1:21
3
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I believe the answer could be:

The second one can be changed to zero as 10 = 2 in Binary!
- One plus zero equals two (in binary).
- Two plus two equals four.
- Five plus ten equals fifteen.
This might be cheating though as I am using '+' as string concatenation.

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  • $\begingroup$ ^vote for lateral thinking. This is the kind of unintended answer I would have to accept as a part of a complete answer. $\endgroup$ – humn Oct 14 '16 at 17:04
2
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Another lateral answer:

Nine of the numbers here could be changed to a specific value so that every statement is true.

three plus three equals six.
one plus four equals five.
six plus ten equals sixteen.

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  • $\begingroup$ ^vote with a note: Those are multiple specific values, not a specific value. $\endgroup$ – humn Oct 14 '16 at 9:49
2
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Following on from @oleslaw's answer:

You could change

The number One in the first statement to any of [2,3,4,5,6,7,8,9,10]

And I believe the statement holds.

I don't think

'Zero of the numbers...' remains true

I have included

10 because we can still change the new value. We can even change it back to One!

As an aside...

I also considered changing the word 'numbers' (one of the 'numbers') to 'words' implying that the second bullet could be changed to 'two times two equals four', but unfortunately I believe the caveat disallows such a funky interpretation. :(

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  • $\begingroup$ ^vote with a note: Would have to see examples to believe this. Am happy to accept unintended answers along with the intended ones as long as they have examples that are explained, and the intended answers should at least be represented by examples. $\endgroup$ – humn Oct 14 '16 at 17:07
1
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Could it be:

One of the numbers here .... None of the numbers here

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  • $\begingroup$ ^vote with a note: Strictly speaking, "zero" would be a better word/value because there is more to the puzzle where that makes a difference. $\endgroup$ – humn Oct 14 '16 at 19:37
1
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Wrap-up: the making of Truth in numbers

This is not a solution to the puzzle but provides notes from its poser. This type of answer has been approved by the community.

Caution: This post contains information about the solution.


Often a new puzzle makes me glad that someone else thought of it because I could never have done so. A new puzzle can also make me wonder why I hadn’t already thought of it. Working out this puzzle was like solving a favorite made by someone else and wishing that I had thought of it.

This began as a casual idea for a quickie but its actual complexity came to light through a series of complete surprises. The original idea was to combine two self-referential paradoxical larks. The first inspiration was Puzzle: digit x appears y times on this piece of paper.

There are ten questions on a piece of paper. Your task is to fill in each blank with a positive integer less than 10 such that there is no contradiction. You can reuse any digit.
•   Digit 0 appears ____ times on this paper.
•   Digit 1 appears ____ times on this paper.
•   Digit 2 appears ____ times on this paper.
•   Digit 3 appears ____ times on this paper.
•   Digit 4 appears ____ times on this paper.
•   Digit 5 appears ____ times on this paper.
•   Digit 6 appears ____ times on this paper.
•   Digit 7 appears ____ times on this paper.
•   Digit 8 appears ____ times on this paper.
•   Digit 9 appears ____ times on this paper.

The second inspiration was an answer to Riddles/puzzles having answer hidden in the question?.

At least two of these statements are false. Ignoring grammar and wordplay, which?
  a. Pigs is pigs.
  b. One plus one is two.
  c. Two plus two does not equal five.
  d. Five and ten make fifteen.

List of false statements:

1. “At least two of these statements are false.”


So it seemed the new puzzle would be answered without much fuss.

 One  Zero of the numbers here could be changed so that every statement is true.
•   One plus one equals two.
•   Two plus two equals four.
•   Five plus ten equals fifteen.

(Click inside hidden areas to permanently reveal them.)

Expected surprise. More answers are likely to be true, such as:

 One Two of the numbers here could be changed so that every statement is true.

Which could lead to:

•   One plus  one  two equals  two  three.
•   Two plus two equals four.
•   Five plus ten equals fifteen.

This unruly multiplicity of possible solutions could be reduced by limiting changes to a single new value, allowing only solutions such as:

 One Three of the numbers here could be changed to a specific value so that every statement is true.
•    One  Zero plus  one  zero equals  two  zero.
•   Two plus two equals four.
•   Five plus ten equals fifteen.

Surprise surprise. The changed first statement makes a new claim, which leads to a third claim, ultimately daisy chaining to a fourth round:

Three of the numbers here could be changed to a specific value so that every statement is true.
•   Zero plus zero equals zero.
•   Zero plus zero equals zero.
•    One  Zero plus  one  zero equals  two  zero.

Blindside broadside. The changed first statement is no longer true! This fourth stage does not contain three numbers that could be changed. Thus each revision, leading all the way back to the first change, actually contained a false statement!   Aww, just one possible answer after all?

Happy surprise ending. Turns out that a valid revised-claim chain can begin with a different number.

 One  Four of the numbers here could be changed to a specific value so that every statement is true.
•   One plus one equals two.
•   Two plus two equals four.
•   Five plus ten equals fifteen.

All statements are indeed true because they can produce:

 One  Four Zero of the numbers here could be changed to a specific value so that every statement is true.
•    One  Zero plus  one  zero equals  two  zero.
•   Two plus two equals four.
•   Five plus ten equals fifteen.

This is ever more interesting than the initial solution as well as its original extensions.

Sure enough, in the spirit of , an unforeseen answer leads to a wonderfully infinite variation, among others.

 1  2 of the numbers here could be changed to a specific value so that every statement is true.
•   1 + 1 = 2.
•   2 + 2 = 4.
•   5 + 10 = 15.

Can lead to:

2 of the numbers here could be changed to a specific value so that every statement is true.
•   1 + 1 = 2.
•   2 + 2 = 4.
•   5 +  1 2 0 =  1 2 5.

Which can remain true by alternating endlessly with:

2 of the numbers here could be changed to a specific value so that every statement is true.
•   1 + 1 = 2.
•   2 + 2 = 4.
•   5 +  2 1 0 =  2 1 5.

Matter of fact, when all values are represented digitally, every number of changes up to 12(!) can be claimed to work.

 1  12 00 of the numbers here could be changed to a specific value so that every statement is true.
•    1 0 +  1 0 =  2 0.
•    2 0 +  2 0 =  4 0.
•    5 0 +  1 0 0 =  15 00.

What a blast to get to naively solve subtly complex abstract puzzles that were never actually thought up by anyone. Some puzzles are created but this one was discovered.

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protected by Aza Oct 19 '16 at 11:08

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