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Mental Math

The largest 3-digit number you can write

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What is the largest number value in base-10 you can write with just 3 digits?

No symbols and characters allowed.

 

Hints: it's not 999

 

Ask someone to write the largest 3-digit number and they'll respond with 999.

 

Logical answer, but we can go bigger.

 

Some may get the "power" brainwave and think of 999 (99 to the power of 9), which calculates out as 99×99×99×99×99×99×99× 99×99.

 

Even better is 999 (9 to the power of 99) which calculates out as 9×9×9×9×9×9×9 ... and so on 99 times.

 

The correct answer, however, if you extend the idea even further ends up as... 99^9 (9 to the 9th power of 9).

Work out the second and third powers first (9×9×9×9×9×9×9×9×9 = 387420489.) We can therefore restate the sum as 9387420489

which works out as.... very very big indeed.

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Semantics: 99^9 is a number with three digits, but I wouldn't consider it a 3-digit number. I wouldn't consider 999 one, either. They contain operators in them.

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yeah, if you're going to allow operators, what's wrong with:

 

[math]9!^{9!^{9!}}[/math] ? that blows [math]9^{9^{9}}[/math]out of the water

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Ok I'll play. I'll use Knuth's up arrow notation to write an expression with three separate single digit numbers. Beat this:

[math]9 \uparrow^{9} \, 9[/math]

Please note the answer is to large for me to display here due to the use of iterated exponentiation (tetration). Of course, we can always beat this result. Does anyone know how?

Use an infinite base such that we have an infinite number of single digit numbers or symbols to represent a number. However a more practicle base would be something like base [math]2^{64}[/math]. That way a single 64 bit number would represent one symbol / digit. We can always choose a larger base. The point being made is that we can always choose a set that has more symbols / elements and push the resulting value of the operation, whichever one you choose, higher.

 

 

Edit - In response to Bignose's post...

yeah, if you're going to allow operators, what's wrong with:

[math]9!^{9!^{9!}}[/math] ? that blows [math]9^{9^{9}}[/math]out of the water

 

I'll modify mine ; )

[math]9! \uparrow^{9!} \, 9![/math]

 

Now that's an operation with an extremely large value that dwarfs [math]9^{9^{9}}[/math] and [math]9!^{9!^{9!}}[/math].

Edited by Daedalus

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If this is a digit here you are:

[math]\infty[/math]

Edited by michel123456

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If this is a digit here you are:

[math]\infty[/math]

Nope that's not a number. It's a limit. If it was a number, I could always add one to it and have a greater resulting value. So far, iterated exponentiation, higher bases, and bignose's factorials gives us the largest number thus far. However, there are operators greater than iterated exponentiation, but I haven't seen any syntax that would represent them. Of course, we could always use infinitely nested factorials...

 

[math]\left(\left(\left(9!\uparrow^{9!}\,9!\right)!\right)!\right)!\,...[/math]

Edited by Daedalus

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Nope that's not a number. It's a limit. If it was a number, I could always add one to it and have a greater resulting value.

it's not a number, but it's maybe a digit. It's an 8 rolled down.

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it's not a number, but it's maybe a digit. It's an 8 rolled down.

 

More like an 8 that got drunk and fell down thinking it was greater than all the other numbers.

Edited by Daedalus

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More like an 8 that got drunk and fell down thinking it was greater than all the other numbers.

yeap.

 

If you can use operator, then you have;

 

1/0

 

------------

in Wikipedia, the [math]\infty[/math] is not in the digits list.

Edited by michel123456

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yeap.

 

If you can use operator, then you have;

 

1/0

 

------------

in Wikipedia, the 'infinite' is not in the digits list.

My post messed up when I went to edit it, but yeah...

However, if we allow infinities then I'll go with:

 

[math]\frac{1}{1-1} = \infty[/math]

 

Then it would be a challenge to see who can get to infinity fastest ; )

Edited by Daedalus

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9!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

and so on, just 1 digit, but rather a big number. However I think 9/(9-9) might be bigger (as long as you ignore the fact that it isn't actually defined)

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9!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

and so on, just 1 digit, but rather a big number. However I think 9/(9-9) might be bigger (as long as you ignore the fact that it isn't actually defined)

That's a rather interesting proposition in the sense that in common notation 9!! = 1*3*5*7*9 < 9!. I don't think I've ever seen more than two exclamation marks, but in a logical continuation you'd probably end up with 9!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! = 1.

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Semantics: 99^9 is a number with three digits, but I wouldn't consider it a 3-digit number

I'm in agreement with this. If you allow that, you might as well allow

a=999

and then perform any operation on a as you would the nines, since a isn't a digit.

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Hmm... Even though allowing operators and such is more of a semantic issue, allowing it gives room for some creative thinking. Consider the sequence of hyper-operations (succession, addition, multiplication, exponentiation, etc.) which continue using right-associativity. Let [math]a\cdot b[/math] denote the 999th such operation of [math]a[/math] by [math]b[/math]. Now take this for size...

 

[math]9\cdot (9\cdot 9)[/math].

 

Of course, anyone can arbitrarily increase any of these parameters or introduce more operators / functions into the expression.

 

Maybe it'd be a bit more challenging to make the restriction to unique operators / functions which are not defined by some boundless parameter. I don't know... factorial, double factorial, prime counting function.

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I take it you mean this

http://en.wikipedia.org/wiki/Double_factorial

which I just looked up.

That's a rather uncommon use of the word common.

Ok,if you insist.

(((((((((9!)!)!)!)!)!)!)!)!)!)!

There are also notions of hyperfactorial and superfactorial, both grow very fast. For example, the superfactorial of 3 has 36 305 decimal digits!

 

One could then define the superduperfactorial by replacing factorial with superfactorial in the definition of the superfactorial. For the number 3, that would have a huge number of decimal digits.

Edited by ajb

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There is an obvious lack of understanding from some members here of what infinite means.

Really? What makes you think that?

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Really? What makes you think that?

The infinite is infinitely larger than any large number.

A large number is less than infinite by an infinite amount.

Edited by michel123456

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We know that. So what?

Are you referring to this?

"If you can use operator, then you have;

1/0"

 

because I have news for you, as I said before, that operator is not defined for zero.

1/0 has no meaningful value (infinite or otherwise)

 

So, there are lots of big numbers in this thread, but the ones which are produced by properly defined use of operators are all finite.

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We know that. So what?

Are you referring to this?

"If you can use operator, then you have;

1/0"

 

because I have news for you, as I said before, that operator is not defined for zero.

1/0 has no meaningful value (infinite or otherwise)

 

So, there are lots of big numbers in this thread, but the ones which are produced by properly defined use of operators are all finite.

 

Exactly... That's why the rules of the game should be to post an expression that uses three single digit base 10 whole numbers to achieve the highest resulting value possible. We should rule out using an operation recursively such as nesting a factorial over and over again, and that only one unary operation can be applied to a number to prevent nesting such number in a sea of unary operations. Nesting a unary operation implies [math]n[/math] recursions, which hides yet another variable or single digit number. For instance, Knuth's up-arrow notation is extended by incorporating a variable that defines such recursion:

 

[math]a \uparrow^{n} b[/math]

 

We can see from the above example that 3 variables are now in play instead of just adding more arrows to the expression.

 

Ajb has introduced super and hyper factorials. So I will introduce exponential factorials and modify my previous result to achieve yet again the largest number to be generated thus far:

 

[math]\text{expofactorial}\left(\text{expofactorial}(9)\uparrow^{\text{expofactorial}(9)}\,\text{expofactorial}(9)\right)[/math]

 

The resulting value is so immense that I do not believe that we could actually calculate the result of the operation without a supercomputer. And even then, your children's children would probably be waiting for the answer evil.gif

Edited by Daedalus

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Might need to be even more precise than that, else I'll go with [math]D(999)[/math] where [math]D:\mathbb{^*R}\to\mathbb{^*R}[/math] is a unary operation that divides its input by the infinitesimal [math]x[/math].

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Might need to be even more precise than that, else I'll go with [math]D(999)[/math] where [math]D:\mathbb{^*R}\to\mathbb{^*R}[/math] is a unary operation that divides its input by the infinitesimal [math]x[/math].

 

Well... if we are ruling out infinities, then I would imagine that we should also rule out infinitesimals for the exact same reason, and only use predefined operators that can be found in Wikipedia, Wolfram, or a website / paper published by an educational institution such as Cornell University's arXiv. Although the operator needs to produce a finite result, it does not necessarily have to be a unary operation given the previous restriction to use three single digit base 10 whole numbers.

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I thought we were only ruling out division by 0, not infinity itself. The set of hyperreals *R includes infinities and infinitesimals, and we can divide a finite number by an infinitesimal to yield infinity. But, no matter.

 

So now the rules are:

 

1. Use three digits and at most one operator (of any convenient arity--and is this one operator per digit, or one operator for the entirety of whatever is constructed by those digits?).

2. We cannot define new operations, though operations defined in the places mentioned in your post are fine.

3. The result must be a real number.

 

Anything I'm missing?

Edited by John

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I'm going to define a function f(x) such that f(x) is the largest possible number that can be expressed using x digits.

Let me know if you can beat f(3).

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