“I just can’t solve this easy puzzle!”

Question

“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?

Answer 1

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!

Answer 2

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.

Answer 3

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).