[Guest]

Using the numbers 1,2,3,4 and the mathematical operations: addition, subtraction, multiplication, division, and POWERS how can you make the largest possible total?

More generally, given the numbers 1...n, how can you make the largest possible total?

Edited June 18, 2010 by psychic_mind

[superprismatic]

Just do a power tower based on n.

[Guest]

I think it is something like this:

2 ^ ( 3 ^ ( 4 + 1 ) )

2 ^ ( 3 ^ ( 4 ^ ( ... ^ ( (n-2) ^ ( (n-1) ^ ( n + 1 ) ) ) ... ) )

Edited June 18, 2010 by lunkkun

[Guest]

You actually get a larger number going reverse

i.e. 3^2 is larger than 2^3, based on this

4^3^(2+1) would be the largest

In general n^(n-1)^(n-2) ... ^4^3^(2+1)

Actually, never mind - I just verified and it's actually bigger starting with the smallest.

Edited June 18, 2010 by littlej

[unreality]

you guys might be interested in

[Guest]

actually, it may b this one:

2^[(3+1)^4] which comes to 2^256.

In gen, 2^{3^[4^...^((n-1)+1)]}^n.

I couldn't check this for just with 5 numbers. Sci-Calc shows Math error!!!!

Edited June 19, 2010 by Praveen S Thivari

[Guest]

For part 2 i don't know the answer so maybe someone would be able to prove their answer as well. I'll have a go later.

Edited June 19, 2010 by psychic_mind

[Guest]

I think it is something like this:

2 ^ ( 3 ^ ( 4 + 1 ) )

if you do 2^(3+1)^4 the number is slightly larger

[Guest]

well I know this isn't gonna win but I've never liked the term 'undefined'

4/(3-2-1) which = 4/0

because in the case of 4/n as 'n' approaches zero then 4/n approaches infinity as 'n' approaches 0.

but in the former case of 4/0 we say it's undefined ... I never did like that term

[Guest]

(1+2)^3^4

[Guest]

Praveen S Thivar!

I don't think anyone can go higher. Unless PVRoot's answer is legal.

To make things easier, let's compare the logarithms.

Since logs are increasing functions for positive numbers, we know that

if log(A) < log(B), then A < B.

I'll write down the expressions shown above, and the corresponding natural logarithms.

2 ^ ( 3 ^ ( 4 + 1 ) ) ::: 168.4348

4^(3^(2+1)) ::: 37.42999

4^3^(2+1) ::: 12.4766

2^[(3+1)^4] ::: 177.4457 *Winner*

(1+2)^(3^4) ::: 88.9876

(1+2)^3^4 ::: 13.1833

Knowing that powers give the best yield, assume the answer is of the form A^B.

We will get better results maximizing B rather than A. So use all numbers except one of them to maximize B. We will use the largest numbers for this. However A cannot be 1, so let A=2.

We now have 1,3,4 to form B, and we follow the same principle.

The only benefit that can be obtained from 1 is through addition.

Where to put the 1 though?

We have 3^4 and we need to insert 1.

Since 3 and 4 are small numbers close together, it is not clear that increasing the position that the 4 is in is better. Let's try both cases.

We end up comparing 4^4 = 256

to 3^5 = 243

So the best answer is 2^((3+1)^4)

Note that the winning answer has 78 digits!

Edited June 26, 2010 by mmiguel1

[Guest]

(1+2)^3^4

(1+2)^34

but I'm not sure if this is permitted.

[Guest]

if the thought process in #12 is legal

then

3⁴²¹ this has 201 digits

and

4³²¹ has 194 digits

and

2⁴³¹ only has 130 digitss

[Guest]

How about (2^(3^(4^(5^(6^(....(n-1^n)+1?

[Guest]

I don't think you can concatenate the numerals.

Parsing the original statement:

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

"Using the numbers 1,2,3,4 and the mathematical operations: addition, subtraction, multiplication, division, and POWERS how can you make the largest possible total?

More generally, given the numbers 1...n, how can you make the largest possible total? "

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

We are given the numbers 1,2,3,4 and are restricted to a set of operations.

You cannot form 12 out of the numbers 1 and 2 with any of the listed operations (without using additional numbers .eg. 1*10 + 2).

You can form the string 12 out of the characters 1 and 2 with concatenation, but by using the word numbers, I believe we are supposed to interpret 1,2,3, and 4 strictly as quantities and not text.

Here is a silly way to interpret strictly as quantities:

Given packages of 1 marble, 2 marbles, 3 marbles, and 4 marbles, and special machines that can combine packages of marbles to produce a new package by the operators listed above, create the package with the most marbles.

You cannot make a package of 12 marbles from the 1 and 2 packages.

[Guest]

If i am not wrong, then ans will be n^((n-1)^((n-2)^...))...

So by calculating this way for 1,2,3,4..,

4^3^2^1= 4^3^2 =4^9 = 262144

[Guest]

Why not:

18249762470488780874564686422801165299572914028994239722316770071597100668834709546023651245269485599114569238294377629242754818885501751993010645278888856753007978697441059800331496768986415104

[Guest]

Three-hundred twenty-one [321] is not one of the numbers we have been given to use. We are given the use of the numbers 1, 2, 3 and 4, not the digits. Nor are We given the concatenation function to use, but only the operations of addition, subtraction, multiplication, division and powers (exponentiation).

The number that you form should be composed of the dydadic operators +, -, *, /, or ^, and, though not mentioned, but inherent to the operations, any brackets needed.

2^((3+1)^4) = 2^(4^4)) = 2^(256) = 115792089237316195423570985008687907853269984665640564039457584007913129639936

Edited July 10, 2010 by Dej Mar