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What is the largest well-defined finite integer that you can exactly describe using unabbreviated English words (no symbols and no letters from other alphabets), with no repeated letters. Any fractions will assume to be rounded down to the nearest integer. Anything that is time-dependent is assumed to refer to the moment the answer is given. The answer does not need to be a grammatically correct sentence so long as it is clear.

By well-defined, I mean a number that can be realistically determined. E.g., 'number of cats' might be large, but can't be determined. However, 'zero' of 'my age in hours' would both be fine (though you should give the answer for the latter).

edit This is not a duplicate. The other question was under-constrained. In fact, I asked this because that question and the answers it got were unsatisfying.

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    $\begingroup$ For whatever it's worth, I agree that this is not only not a duplicate but also an infinitely better and more interesting question than the one it's being compared to. I could elaborate but I'm already bored by the potential ensuing argument. $\endgroup$ – question_asker Dec 7 '15 at 19:01
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    $\begingroup$ Did you mean "'zero' or 'my age in hours' ..."? Zero of an age is somewhat vague. $\endgroup$ – Lawrence Dec 12 '15 at 12:09
  • $\begingroup$ Please do not edit your question in a way that fundamentally change it. In particular, do not edit your question in a way that invalidates existing answers. If you'd like to ask your edited question, with the new constraint that eliminates answers that refers to other answers, ask it as a new answer. $\endgroup$ – Gilles Dec 12 '15 at 14:05
  • $\begingroup$ @Gilles Surely you don't expect everyone to always foresee every possible almost-funny-the-first-time-you-see-it loophole joke. I think the asker (here and generally) has a right to disqualify boring "cheating" answers. We even have two Meta discussions about this very thing: meta.puzzling.stackexchange.com/questions/1334/… meta.puzzling.stackexchange.com/questions/1302/… $\endgroup$ – question_asker Dec 12 '15 at 14:18
  • $\begingroup$ @question_asker Refering to other answers isn't a loophole here. It's explicitly allowed by the question: the question allows refering to real-world objects — and this thread is a real-world object. The meta threads you refer to are about disqualifying answers that violate explicit assumptions of the question, in particular questions being a mathematical puzzle rather than a word puzzle or situational puzzle. The question here is a word puzzle. $\endgroup$ – Gilles Dec 12 '15 at 14:29
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Edit: a much larger answer.

Two up hymn age six

Which I will interpret as 2, followed by (hymn age) of Knuth up-arrows, six. The age of hymns (according to wikipedia, and we should probably go with the western variety) is roughly 2,650 [years, the default English interpretation of age], thanks to the ancient Greeks.

This gives us:

2↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑6

Which is less than Graham's number (in fact, less than g2, where Graham's is g64), but larger than anything else you could possibly conceive without also using up-arrow notation.


Let's start this off with something decent:

Planck time hour

An hour in units of "Planck time" (per Wolfram Alpha) is roughly

$6.7 \times 10^{46}$

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Alternatively, combining two of the other ideas in here:

Sixty Mole bang

Which is $60(6.022141 \times 10^{23})!$, or (again, thanks Wolfram Alpha) roughly

$10^{10^{25}} = 1 \times 10^{10000000000000000000000000}$

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But increasingly tortured readings of English may allow us to improve this to:

Mole sixty bang

which is $(6.022141 \times 10^{23})!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!$, or roughly

$1 \times 10^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{10}^{{25}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

I'd suggest some up-arrow notation at this point...

$(10↑↑61)^{25}$

Of course, you might then also allow for

Sixty Mole bang

where the mole applies to the number of bangs. This gets messy quickly, but a very rough estimate would be:

$10↑↑(10↑24)$

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  • $\begingroup$ This wins it if nobody comes up with something better before a week is out. I'm playing around with 'mole' which gets us to 6.0*10^23 with letters to spare. It's unitless so really is just a number. $\endgroup$ – Dave Dec 9 '15 at 14:07
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    $\begingroup$ Are you sure this isn't just a porn title? :D $\endgroup$ – Irishpanda Dec 9 '15 at 17:04
  • $\begingroup$ Since you're not using brackets, order of operations dictates that this be interpreted as 60*[(6.02*10^23)!], which is substantially less, though still ludicrously large. $\endgroup$ – Ninety-Three Dec 9 '15 at 17:14
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    $\begingroup$ Well, sort of. A Mole of X is Avagadro's Number of X, so using Mole as a noun equivalent to Avagadro's Number is certainly poor grammar. But, conveniently, we're not required to be grammatically correct - just clear and determinable. I think Mole is good enough for those purposes. $\endgroup$ – Zerris Dec 9 '15 at 19:37
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    $\begingroup$ I suppose it all comes down to the minutia of what the question means by "clear". In the context of up arrows, it's fairly clear (it's the most logical interpretation anyway), but I doubt most people would go to Knuth without already knowing the context. $\endgroup$ – Ninety-Three Dec 10 '15 at 5:24
5
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Enormabixul

The Googology wiki has many like these. This one is many, many orders of magnitude greater than, for example, Graham's number.

The Rayo sum

Rayo's number is one of the greatest numbers ever defined in mathematics. From wikipedia:

The smallest number bigger than any finite number named by an expression in the language of set theory with a googol symbols or less.

Note that this definition is not self-referential, since it can't be expressed using only (first-order) set theory.

A general list of googologisms can be found in the wiki. There is also a list of fast-growing functions.

This may be useful for future answerers.

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3
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Top squared.

Whichever answer is the top one (either Ivan Barreto's or Zerris's, depending on whether you consider enormabixul an English word — Rayo isn't, it's only a proper name), I'm squaring that.

Since answers are to be interpreted at the time of posting, I'm not closing the field.

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  • $\begingroup$ ...I like it. But there's just one problem - either your answer is the largest, or it isn't. If it isn't the largest, we clearly shouldn't accept it as the answer to this question. If it is the largest, then it is simply asking to square itself, and it provides no numeric value - ergo, it's nonsense, and it being the largest is contradictory. Since one possibility is a contradiction and the other is not, we must be in the non-contradictory state, and this answer must not be the largest. You could probably get around this with slightly cleverer less-direct phrasing. $\endgroup$ – Zerris Dec 12 '15 at 9:24
  • $\begingroup$ @Zerris Answers are to be interpreted at the time of posting, as per the question (the question at the time I posted my answer, I mean, I notice that it's since been edited). At the time I posted my answer, my answer was not present, so it could not have been the top answer. $\endgroup$ – Gilles Dec 12 '15 at 14:04
  • $\begingroup$ Ah, right you are - in which case, yes, such answers are (were) valid but inherently break the question format, since each person can respond in kind. $\endgroup$ – Zerris Dec 12 '15 at 16:51
3
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My best and four

Just to show that the list can go on.

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0
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How about:

'MDCLXVI score bang'

Using the translation

MDCLXVI = 1666 in Roman numerals,
score = 20 in 'Old' English, and
bang = equivalent to the factorial symbol

I get $(1666 \cdot 20)! = 33320! \approx 3.1525677 \cdot10^{136225}$

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    $\begingroup$ I'll argue that "MDCLXVI" is not an English word. $\endgroup$ – Zerris Dec 7 '15 at 2:54
  • $\begingroup$ You have a point, but I would argue that Roman Numerals have been borrowed into English in the same manner as foreign language terms such as "Pinot Noir". $\endgroup$ – Penguino Dec 7 '15 at 3:22
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    $\begingroup$ For example, it is commonly used in the shorthand for a monarch' or Pope's name (Henry V for Henry the fifth, Benedict XVI etc.), and in the credits of a movie to denote the year of production (and in books to denote year of publication). $\endgroup$ – Penguino Dec 7 '15 at 3:30
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    $\begingroup$ That would, however, make it symbols rather than words - equivalent to just using the numbers themselves. $\endgroup$ – Zerris Dec 7 '15 at 3:34
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    $\begingroup$ You are using 'C' twice $\endgroup$ – The Dark Truth Dec 7 '15 at 9:43
-1
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g sub forty k

That's $g_{40000}$, where a lowercase $g$ is Graham's sequence.

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  • $\begingroup$ Neither g nor k are English words. $\endgroup$ – Zerris Dec 10 '15 at 16:13
  • $\begingroup$ You could also say Forcal, which is g sub one million, but is still much smaller than my entry, enormabixul :) $\endgroup$ – MathET Dec 10 '15 at 16:21
  • $\begingroup$ @Zerris: Only other languages' letters are disallowed by the question. $\endgroup$ – Deusovi Dec 10 '15 at 16:22
-1
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I'm gonna play off Zerris' answer and go with

Foe Sixty Bang

https://en.wikipedia.org/wiki/Foe_(unit)

A foe is a unit of energy equal to 10^44 joules or 10^51 ergs, used to measure the large amount of energy released by a supernova.

So,

$(1 \times 10^{44})!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!$

or better yet:

$(1 \times 10^{51})!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!$ in ergs

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    $\begingroup$ I think unlike mole (which is a unitless number), this has a unit, so it's 1 foe unless you can say something like 'foe jules' or 'foe ergs' or something along those lines. $\endgroup$ – Dave Dec 9 '15 at 23:45
-4
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"The largest number ever used in a serious mathematical proof."

Wich apparently isn't Graham's Number anymore:

Specific integers known to be far larger than Graham's number have since appeared in many serious mathematical proofs.

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    $\begingroup$ You aren't allowed to use letters more than once. And welcome to Puzzling SE. $\endgroup$ – The Dark Truth Dec 7 '15 at 11:06
  • $\begingroup$ Ah, indeed. Overlooked that detail. Too bad even "Graham's Number" has duplicate letters... $\endgroup$ – Zuzel Dec 7 '15 at 11:22

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