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“Yuh can’t solve this easy puzzle?”
                   “Bah, it’s already solved!”
“Nah, it’s so easy it only looks solved.”

$ \require{begingroup}\begingroup \def \a #1#2#3{ \abc {three}{#1}{nine} {#2}{thirteen}{#3} } \def \b #1#2#3{ &&\abc {five}{#1}{seven} {#2}{fourteen}{#3} } \def \c #1#2#3{ && \abc{eight}{#1}{eleven}{#2}{seventeen}{#3} } \def \abc #1#2#3#4#5#6{ \phantom {\sf#1} \llap{\sf#2} {\small\, +\, } \rlap{\sf#4} \phantom{\sf#3} { \small\,= \;} \rlap {\sf\, #6\, } \rlap{\underline{\hphantom { \sf\, #6 \, }}} \hphantom{ \underline{ \sf\,#5 \,}} } \bbox[darkgreen,5pt]{\color{lightyellow}{\begin{matrix} \raise.5ex\strut \a{ one}{ten }{eleven } \b{ six}{seven}{thirteen} \c{ six}{eleven}{seventeen} \\ \a{ two}{ten }{twelve } \b{five}{nine }{fourteen} \c{eight}{nine }{seventeen} \\ \a{ six}{six }{twelve } \b{ six}{eight}{fourteen} \c{eight}{ten }{eighteen } \\ \a{three}{ten }{thirteen} \b{five}{ten }{fifteen } \c{ nine}{nine }{eighteen } \\ \a{ four}{nine}{thirteen} \b{ six}{nine }{fifteen } \c{ ten}{ten }{twenty } \raise-.5ex\strut \end{matrix}}} \endgroup $

“Ah, and it isn’t even completely posed.”
                    “?”
“Uh, it’s missing an entry.”
                    “Oh?   Ohh.   Yeahhh.   Too easy.”

          What entry is missing?

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  • 2
    $\begingroup$ I know the pattern but I can't find another that fits for my life :P $\endgroup$ – Beastly Gerbil Apr 23 at 21:57
  • 1
    $\begingroup$ @BeastlyGerbil You and me both! :) $\endgroup$ – Stiv Apr 23 at 22:00
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    $\begingroup$ All I know is I'll probably be kicking myself when someone gets it :P. I don't think I'm meant for this lol $\endgroup$ – Beastly Gerbil Apr 23 at 22:07
  • 1
    $\begingroup$ In light of @Stiv's answer, the font choice (where the summands are set in upright type, and the sum is set in badly-spaced slanted type (as a result of being in math mode)) is deceptive, I think. Would you be amenable to putting all the words in upright type? $\endgroup$ – LSpice Apr 26 at 22:57
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    $\begingroup$ I think that looks way better! $\endgroup$ – LSpice Apr 27 at 4:19
44
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In this puzzle, all of the sums:

Involve numbers between 1 and 20* (inclusive), where the total number of letters in the numbers being added together equals the number of letters in their resulting sum.

For example, 'one + ten = eleven' is involved in this list because 'one' has 3 letters, 'ten' has 3 letters, and 'eleven' has 3+3=6 letters.

*Note that the '1 to 20' criterion avoids the need to enter arguments about whether to consider numbers whose names are spelled out using 'more than one word' (requiring either spaces or hyphenation). Remember, we are seeking a unique entry that can be added to this set (as the OP's use of 'an entry' implies that we seek only one answer). If we assume no number greater than 20 can be involved, we can totally avoid the need to debate the many potential 'multi-word' equations like 'ten + eleven = twenty one' or 'ten + twelve = twenty two' (wherein both sides have 9 letters).

However, try as I might (and having applied some pretty strict criteria) I just could not find another pair of numbers which could belong to this set - no matter which new combination of two numbers I tested, it seemed that the set was already totally complete! I debated loosening my criteria, widening my search... And then it finally hit me!

The missing member of the set isn't a sum involving a pair of numbers - it involves a triple!

This way, the entry which is uniquely missing from this list is:

$$one + six + ten = seventeen$$
...as each side of this equation has 9 letters in total. After all, who said anything about the left-hand side having to have just 2 terms??!

Very clever - had me thinking for a loooong time there!

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  • $\begingroup$ Ahhhhhh thats what I was missing :P +1 $\endgroup$ – Beastly Gerbil Apr 23 at 22:27
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    $\begingroup$ @BeastlyGerbil I cannot put into words how much I had to rack my brains for this one! I went through the list SO MANY TIMES. humn: That was DEVIOUS - well played :) $\endgroup$ – Stiv Apr 23 at 22:28
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    $\begingroup$ Not exactly 1 to 10; look at the first entry in the third column. But otherwise +1 $\endgroup$ – shoover Apr 24 at 16:09
  • $\begingroup$ @shoover Oops, typo - thanks! :) $\endgroup$ – Stiv Apr 24 at 16:25
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    $\begingroup$ Twenty-one is a single word. $\endgroup$ – Nij Apr 25 at 3:23
7
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I assume it's for positive numbers that less and equal to 20. (Since twenty one has space). Here is Python code for solving this question:

At first, I thought it uses two numbers only:

text = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen", "twenty"]
number = list(range(1,21))

result = set()
for i in range(len(text)):
    for j in range(len(text)):
        num1 = number[i]
        num2 = number[j]
        summ = num1 + num2
        if summ in number and len(text[i]) + len(text[j]) == len(text[summ-1]):
            if text[i] < text[j]:
                res = f"{text[i]} + {text[j]} = {text[summ-1]}"
            else:
                res = f"{text[j]} + {text[i]} = {text[summ-1]}"
            result.add(res)
for i in result:
    print(i)

But the questions already includes all two numbers, so I try three numbers:

text = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen", "twenty"]
number = list(range(1,21))

result = set()
for i in range(len(text)):
    for j in range(len(text)):
        for k in range(len(text)):
            num1 = number[i]
            num2 = number[j]
            num3 = number[k]
            summ = num1 + num2 + num3
            if summ in number and len(text[i]) + len(text[j]) + len(text[k]) == len(text[summ-1]):
                t = [text[i], text[j], text[k]]
                t.sort()
                res = " + ".join(t) + " = " + text[summ-1]
                result.add(res)
for i in result:
    print(i)

And got the answer:

one + six + ten = seventeen

And:

There won't be solutions for more than 3, since minimum length is 3, 4 numbers minimum length is 12, which is longer than any numbers.

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  • $\begingroup$ I absolutely love your Python answers! I'm learning Python myself and its great to see applicability! $\endgroup$ – ThatOneNerdyBoy Apr 26 at 18:18
4
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Based on the rule as explained in the other answer

the sums work not only as numbers but also as wordlengths

— other possible equations are

fifty+fifty=one hundred

and

sixty+forty=one hundred

(not to mention numerous trivial examples along the lines of

three thousand two+twenty=three thousand twenty-two,

which I guess don't count).

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  • 2
    $\begingroup$ Uniqueness is part of the criterion for this puzzle. How would you include either one of those answers while preserving uniqueness? $\endgroup$ – Galen Apr 24 at 0:14
  • $\begingroup$ @Galen, then I guess the puzzle was ill-contrived: If the rule is indeed a stated in the other answer, then there isn't uniqueness. $\endgroup$ – msh210 Apr 24 at 0:20
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    $\begingroup$ Yeah, maybe so. Stiv constrains their answer to summands between and including 1 and 10, but I can see how it is pretty trivial for create unions of other intervals that give one answer. This puzzle could use some refinement. $\endgroup$ – Galen Apr 24 at 0:27
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    $\begingroup$ I think the rule is only numbers which are singular words are used. This would rule out all other possibilities $\endgroup$ – Beastly Gerbil Apr 24 at 0:32
  • 1
    $\begingroup$ @humn Thanks! Done. $\endgroup$ – msh210 Apr 24 at 6:56

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